Question 1

$8^{2.4} 	imes 2^{3.7} \div (16)^{1.3} = 2^{x}$

Accepted Answer

Concept & Formula: This problem requires applying the Laws of Indices to simplify an expression by finding a common prime base. The essential exponent rules are: (a^m)^n = a^{m × n} a^m × a^n \div a^p = a^{m+n-p} Step-by-Step Solution: • Step 1: Identify the lowest common base, which is 2. Convert all other bases (8 and 16) to powers of 2. 8^{2.4} = (2^3)^{2.4} = 2^{7.2} (16)^{1.3} = (2^4)^{1.3} = 2^{5.2} • Step 2: Substitute these converted values back into the original equation. 2^{7.2} × 2^{3.7} \div 2^{5.2} = 2^x • Step 3: Apply the multiplicative and divisional laws of indices to combine the exponents on the Left Hand Side (LHS). 2^{7.2 + 3.7 - 5.2} = 2^x 2^{10.9 - 5.2} = 2^x 2^{5.7} = 2^x • Step 4: Equate the powers since the bases on both sides are identical. x = 5.7 Exam Strategy & Shortcut: To solve this rapidly, skip writing down the intermediate base steps. Mentally multiply the given powers by their base's relation to 2 (e.g., multiply 8's power by 3, and 16's power by 4). Just write down the final exponent string: 3(2.4) + 3.7 - 4(1.3) = 7.2 + 3.7 - 5.2. This avoids rewriting the base 2 repeatedly and directly yields 5.7. Common Pitfall: Students often mistake the division sign for a subtraction sign or misapply the division rule by dividing the exponents instead of subtracting them. Always remember that a^m \div a^n means subtracting the exponent n from m. Final Answer: Therefore, the correct answer is 5.7.

Question 2

$(64x^3 \div 27a^{-3})^{-\frac{2}{3}}$

Accepted Answer

Concept & Formula: This problem tests your mastery of the Laws of Indices, specifically dealing with negative exponents and fractional powers. The key exponent rules required are: a^{-m} = (1/a^m) ((a/b))^{-m} = ((b/a))^m (a^m)^n = a^{m × n} Step-by-Step Solution: • Step 1: Simplify the expression inside the parentheses by removing the negative exponent on the variable a. 64x^3 \div 27a^{-3} = (64x^3/27a^{-3)} Since a^{-3} is in the denominator, it moves to the numerator as a^3: (64x^3 a^3/27) • Step 2: Apply the negative fractional exponent -(2/3) to the entire fraction. A negative exponent flips the fraction. ((64x^3 a^3/27))^{-(2/3)} = ((27/64x^3 a^3))^{(2/3)} • Step 3: Evaluate the fractional power (2/3), which means taking the cube root (denominator 3) and then squaring the result (numerator 2). Let's take the cube root of the numerator and denominator first. \sqrt[3]{27} = 3 \sqrt[3]{64x^3 a^3} = 4xa So, the expression becomes: ((3/4xa))^2 • Step 4: Square both the numerator and the denominator. (3^2/(4xa)^2) = (9/16x^2 a^2) Exam Strategy & Shortcut: To solve this rapidly, handle the constants and variables separately. For the constants: (64 \div 27)^{-2/3}. Flip it to get (27/64)^{2/3}. The cube root is 3/4, and squaring it gives 9/16. This eliminates options (a) and (b). For the variables: x^3 and a^{-3} in the denominator becomes x^3 a^3. Raising (x^3 a^3) to the power of -2/3 directly gives x^{-2} a^{-2}, which means they both belong in the denominator squared. This immediately points to option (c). Common Pitfall: A frequent error is forgetting to apply the exponent to the coefficients (64 and 27) or incorrectly flipping the fraction when dealing with the negative fractional exponent, leading to inverted answers like (16/9). Final Answer: Therefore, the correct answer is (9/16x^2 a^2).

Question 3

The value of $\frac{2^{3x+4} + 8^{x+1}}{8^{x+1} - 2^{3x+2}}$ is

Accepted Answer

Concept & Formula: This question requires converting different bases to a common prime base (2 in this case) and subsequently factoring out terms to simplify the fraction. The governing exponent rule is: (a^m)^n = a^{m × n} Step-by-Step Solution: • Step 1: Convert the base 8 terms to base 2. We know 8 = 2^3. 8^{x+1} = (2^3)^{x+1} = 2^{3(x+1)} = 2^{3x+3} • Step 2: Substitute this converted term back into the original fraction. Numerator: 2^{3x+4} + 2^{3x+3} Denominator: 2^{3x+3} - 2^{3x+2} • Step 3: Factor out the smallest power of 2 from both the numerator and the denominator. Numerator: Factor out 2^{3x+3} 2^{3x+3} (2^1 + 1) = 2^{3x+3} (3) Denominator: Factor out 2^{3x+2} 2^{3x+2} (2^1 - 1) = 2^{3x+2} (1) • Step 4: Reconstruct the fraction with the factored expressions. \frac{2^{3x+3} × 3}{2^{3x+2}} • Step 5: Simplify the exponent term using the division law. 2^{(3x+3) - (3x+2)} × 3 2^1 × 3 = 6 Exam Strategy & Shortcut: Variable Substitution: Since the options are all constants, the expression's value is independent of x. Choose the simplest value for x, which is x = 0. Substitute x = 0: Numerator: 2^4 + 8^1 = 16 + 8 = 24 Denominator: 8^1 - 2^2 = 8 - 4 = 4 Result: (24/4) = 6. This method resolves the problem in roughly 10 seconds without needing to apply index laws. Common Pitfall: A common mistake when applying the exponent rule (a^m)^n is failing to distribute the multiplier across the entire binomial. Students might write (2^3)^{x+1} as 2^{3x+1} instead of the correct 2^{3x+3}, which cascades into a completely wrong result. Final Answer: Therefore, the correct answer is 6.

Question 4

The value of $\frac{2^{n-1} - 2^n}{2^{n+4} + 2^{n+1}}$ is

Accepted Answer

Concept & Formula: The core concept is factoring exponential expressions by extracting the common term with the lowest power to simplify the numerator and denominator separately. The primary rule utilized is: a^{m+n} = a^m × a^n Step-by-Step Solution: • Step 1: Isolate the lowest power of 2 in both the numerator and the denominator to factor them out. In the numerator (2^{n-1} - 2^n), the lowest power is n-1. Note that 2^n can be written as 2^{n-1} × 2^1. In the denominator (2^{n+4} + 2^{n+1}), the lowest power is n+1. Note that 2^{n+4} can be written as 2^{n+1} × 2^3. • Step 2: Factor out these lowest power terms. Numerator: 2^{n-1} (1 - 2^1) = 2^{n-1} (1 - 2) = 2^{n-1} (-1) = -2^{n-1} Denominator: 2^{n+1} (2^3 + 1) = 2^{n+1} (8 + 1) = 2^{n+1} (9) • Step 3: Reconstruct the fraction with the simplified terms. \frac{-2^{n-1}}{9 × 2^{n+1}} • Step 4: Simplify the exponent parts by bringing 2^{n-1} down to the denominator using the division law a^m \div a^n = a^{m-n}. (-1/9 × 2^{(n+1) - (n-1))} (-1/9 × 2^{n+1-n+1)} (-1/9 × 2^2) • Step 5: Calculate the final arithmetic value. (-1/9 × 4) = -(1/36) Exam Strategy & Shortcut: Assume n=1: Since the options are all distinct numerical constants, the outcome is independent of the variable n. Choosing a small value like n=1 makes the calculation trivial. Numerator: 2^0 - 2^1 = 1 - 2 = -1 Denominator: 2^5 + 2^2 = 32 + 4 = 36 Result: -(1/36). This takes only moments to execute mentally. Common Pitfall: A very common algebraic mistake here is mishandling the subtraction in the numerator (1 - 2). Students frequently drop the negative sign by mistake and answer (1/36), which isn't an option here but could easily trap someone if it were. Final Answer: Therefore, the correct answer is -(1/36).

Question 5

$\sqrt{6 - 4\sqrt{3} + \sqrt{16 - 8\sqrt{3}}}$ is equal to

Accepted Answer

Concept & Formula This problem involves nested square roots. The strategy is to simplify from the innermost square root outwards by turning binomial surds into perfect squares using the identity: (a - b)^2 = a^2 + b^2 - 2ab Step-by-Step Solution • Start with the innermost root: \sqrt{16 - 8\sqrt{3}} • Convert the term -8\sqrt{3} to the form -2ab. 16 - 8\sqrt{3} = 16 - 2(4\sqrt{3}) = 16 - 2\sqrt{16 × 3} = 16 - 2\sqrt{48} • We need two numbers that add to 16 and multiply to 48. These are 12 and 4. • Rewrite: (\sqrt{12})^2 + (\sqrt{4})^2 - 2\sqrt{12}\sqrt{4} = (\sqrt{12} - \sqrt{4})^2 • Simplify \sqrt{12} - \sqrt{4} to 2\sqrt{3} - 2. • So, the innermost root evaluates to 2\sqrt{3} - 2. (Note: 2\sqrt{3} > 2, so it is positive). • Substitute this back into the main expression: \sqrt{6 - 4\sqrt{3} + (2\sqrt{3} - 2)} • Combine like terms: \sqrt{4 - 2\sqrt{3}} • Now, simplify this new root. We need two numbers adding to 4 and multiplying to 3. They are 3 and 1. 4 - 2\sqrt{3} = 3 + 1 - 2\sqrt{3} = (\sqrt{3} - 1)^2 • Taking the square root gives \sqrt{3} - 1. Exam Strategy & Shortcut For nested surds, always work from the inside out. When simplifying 16 - 8\sqrt{3}, instead of pushing the whole 4 into the root to make \sqrt{48}, you can mentally break down 8\sqrt{3} as 2 × 4 × \sqrt{3}. Ask yourself: Does 4^2 + (\sqrt{3})^2 = 16? Yes! So it's immediately (4 - \sqrt{3})^2. No, wait. 16 + 3 = 19, not 16. Let's break it differently: 2 × (2\sqrt{3}) × 2. Does (2\sqrt{3})^2 + 2^2 = 16? 12 + 4 = 16. Yes! So the root is 2\sqrt{3} - 2. This mental check is much faster than full expansion. Common Pitfall When taking the square root of (x - y)^2, always ensure x > y so the result is positive. For instance, if you incorrectly write \sqrt{4 - 2\sqrt{3}} as 1 - \sqrt{3}, you will get a negative value, which violates the principle of real square roots. Final Answer Therefore, the correct answer is \sqrt{3} - 1.

Question 6

The value of $\frac{1}{\sqrt{12 - \sqrt{140}}} - \frac{1}{\sqrt{8 - \sqrt{60}}} - \frac{2}{\sqrt{10 + \sqrt{84}}}$ is

Accepted Answer

Concept & Formula To simplify a surd of the form \sqrt{a \pm 2\sqrt{b}}, we find two numbers whose sum is a and product is b. The expression then becomes a perfect square (\sqrt{x} \pm \sqrt{y})^2. If the 2 is missing before the inner root, extract it from inside the root by dividing the inner number by 4. (x \pm y)^2 = x^2 + y^2 \pm 2xy Step-by-Step Solution • Evaluate the first denominator: \sqrt{12 - \sqrt{140}} Extract a 4 from 140 to get a 2 outside: \sqrt{12 - 2\sqrt{35}} We need numbers that add to 12 and multiply to 35. They are 7 and 5. So, \sqrt{12 - 2\sqrt{35}} = \sqrt{7} - \sqrt{5} • Evaluate the second denominator: \sqrt{8 - \sqrt{60}} Extract a 4 from 60: \sqrt{8 - 2\sqrt{15}} Numbers adding to 8 and multiplying to 15 are 5 and 3. So, \sqrt{8 - 2\sqrt{15}} = \sqrt{5} - \sqrt{3} • Evaluate the third denominator: \sqrt{10 + \sqrt{84}} Extract a 4 from 84: \sqrt{10 + 2\sqrt{21}} Numbers adding to 10 and multiplying to 21 are 7 and 3. So, \sqrt{10 + 2\sqrt{21}} = \sqrt{7} + \sqrt{3} • Rationalize each term in the main expression: Term 1: (1/\sqrt{7) - \sqrt{5}} = \frac{\sqrt{7} + \sqrt{5}}{7 - 5} = \frac{\sqrt{7} + \sqrt{5}}{2} Term 2: (1/\sqrt{5) - \sqrt{3}} = \frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \frac{\sqrt{5} + \sqrt{3}}{2} Term 3: (2/\sqrt{7) + \sqrt{3}} = \frac{2(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{2(\sqrt{7} - \sqrt{3})}{4} = \frac{\sqrt{7} - \sqrt{3}}{2} • Combine all rationalized terms: \frac{\sqrt{7} + \sqrt{5}}{2} - \frac{\sqrt{5} + \sqrt{3}}{2} - \frac{\sqrt{7} - \sqrt{3}}{2} = \frac{\sqrt{7} + \sqrt{5} - \sqrt{5} - \sqrt{3} - \sqrt{7} + \sqrt{3}}{2} = (0/2) = 0 Exam Strategy & Shortcut Instead of writing out the full rationalization step, recognize the pattern: (1/\sqrt{a) - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}. When you identify the differences between the numbers inside the roots (e.g., 7 - 5 = 2, 5 - 3 = 2), you can immediately put them all over a common denominator of 2 and cancel the alternating positive and negative terms mentally to save crucial seconds. Common Pitfall Failing to distribute the negative signs correctly when combining the fractions. Writing -\frac{\sqrt{5} + \sqrt{3}}{2} and accidentally calculating it as -\sqrt{5} + \sqrt{3} instead of -\sqrt{5} - \sqrt{3} is a frequent algebraic mistake that leads to non-zero answers. Final Answer Therefore, the correct answer is 0.

Question 7

The value of $$ \left(x^{\frac{b+c}{c-a}}\right)^{\frac{1}{a-b}} \cdot \left(x^{\frac{c+a}{a-b}}\right)^{\frac{1}{b-c}} \cdot \left(x^{\frac{a+b}{b-c}}\right)^{\frac{1}{c-a}} $$ is

Accepted Answer

Concept & Formula: When raising a power to a power, multiply the exponents: (x^m)^n = x^{mn} . When multiplying terms with the same base, add their exponents: x^m \cdot x^n = x^{m+n} . For cyclic symmetry problems in competitive exams, the sum of the exponents almost always evaluates to zero. Step-by-Step Solution: • Step 1: Apply the power-to-a-power rule to simplify each term's exponent. First term exponent = (b+c/(c-a)(a-b)) Second term exponent = (c+a/(a-b)(b-c)) Third term exponent = (a+b/(b-c)(c-a)) • Step 2: Multiply the bases by adding all three exponents together. Total Exponent = (b+c/(c-a)(a-b)) + (c+a/(a-b)(b-c)) + (a+b/(b-c)(c-a)) • Step 3: Take the Least Common Multiple (LCM) for the denominators, which is (a-b)(b-c)(c-a) . Adjust the numerators accordingly: First term numerator = (b+c)(b-c) = b^2 - c^2 Second term numerator = (c+a)(c-a) = c^2 - a^2 Third term numerator = (a+b)(a-b) = a^2 - b^2 • Step 4: Sum the numerators. (b^2 - c^2) + (c^2 - a^2) + (a^2 - b^2) = 0 • Step 5: Evaluate the final base and exponent. Since the numerator of our exponent is 0, the total exponent is 0. x^0 = 1 (assuming x 
eq 0) Exam Strategy & Shortcut: Pattern Recognition: Cyclic equations (where variables shift in a continuous loop like a ightarrow b ightarrow c) usually simplify symmetrically. In exponent problems of this exact structure, the exponents invariably cancel out to yield x^0 = 1. Value Substitution: If you forget the algebra, assume a=1, b=2, c=3. (Never pick values that make a denominator zero). The arithmetic will quickly cancel down to x^0 = 1, allowing you to mark the answer in under 15 seconds. Common Pitfall: A major trap is making sign errors when establishing the common denominator. For example, treating (c-a) as -(a-c) and losing track of the negative sign will prevent the numerators from perfectly canceling out. Final Answer: Therefore, the correct answer is 1.

Question 8

If $2^x = 4^y = 8^z$ and $$ \left(\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z}\right) = \frac{24}{7} $$ then the value of $z$ is

Accepted Answer

Concept & Formula: When given a chain of exponential equalities with bases that are powers of the same number (e.g., 2, 4, 8), convert all bases to the smallest common prime base. Then, express all variables in terms of the specific variable you need to find. Step-by-Step Solution: • Step 1: Express the exponential chain using a common base of 2. 2^x = (2^2)^y = (2^3)^z 2^x = 2^{2y} = 2^{3z} • Step 2: Equate the exponents to find the relationship between x, y, and z. x = 2y = 3z • Step 3: The problem asks for the value of z. Therefore, convert all terms in the given fractional equation into terms of z. We know: x = 3z \rightarrow 2x = 6z 2y = 3z \rightarrow Multiply by 2 \rightarrow 4y = 6z • Step 4: Substitute these z-based values into the given equation. (1/2x) + (1/4y) + (1/6z) = (24/7) (1/6z) + (1/6z) + (1/6z) = (24/7) • Step 5: Simplify and solve for z. (3/6z) = (24/7) (1/2z) = (24/7) Cross-multiply: 7 = 48z z = (7/48) Exam Strategy & Shortcut: Notice the symmetry in the fractional equation once you find x = 2y = 3z. The denominators 2x, 4y, and 6z look different, but checking their base relationships (2(3z) = 6z, 2(2(2y)) \dots wait, 2y = 3z \rightarrow 4y = 6z) reveals they are completely identical. Recognizing instantly that the equation is just 3 × (one term) = (24/7) allows you to jump straight to (1/2z) = (24/7), solving the problem mentally in under 30 seconds. Common Pitfall: A frequent mistake is solving for the wrong variable (like finding x instead of z) and then scrambling to convert it at the end, which introduces fraction arithmetic errors. Always substitute so that the only remaining variable is the one requested by the question. Final Answer: Therefore, the correct answer is (7/48).

Question 9

If $3^{(x - y)} = 27$ and $3^{(x + y)} = 243$, then $x$ is equal to

Accepted Answer

Concept & Logic: To solve exponential equations, express both sides of the equation with the exact same base. Once the bases match, you can directly equate their exponents to form a system of linear equations. Step-by-Step Solution: • Step 1: Express the numbers on the right side of both equations as powers of the base 3. We know that 27 = 3 × 3 × 3 = 3^3. We know that 243 = 27 × 9 = 3^3 × 3^2 = 3^5. • Step 2: Rewrite the original equations with matching bases. Equation 1: 3^{(x - y)} = 3^3 Equation 2: 3^{(x + y)} = 3^5 • Step 3: Equate the exponents, as the bases (3) are equal. (1) x - y = 3 (2) x + y = 5 • Step 4: Solve the resulting system of linear equations. The easiest method here is addition. Add equation (1) and equation (2) together: (x - y) + (x + y) = 3 + 5 2x = 8 • Step 5: Isolate x. x = (8/2) x = 4 Exam Strategy & Shortcut: Mental Math Elimination: The powers of 3 are standard knowledge (3, 9, 27, 81, 243). You can instantly read the equations as "sum is 5, difference is 3". What two numbers add to 5 and subtract to 3? They must be 4 and 1. Since the question asks for x (the larger number, since x-y is positive), x must be 4. This bypasses formal written steps entirely. Common Pitfall: A common error is solving for the wrong variable in the rush of the exam. After setting up the equations, a student might subtract them to find 2y = 2 \rightarrow y = 1, and then accidentally select an option matching y if one were provided. Always double-check which variable the question is asking for. Final Answer: Therefore, the correct answer is 4.

Question 10

If $2^{2x-1} + 4^x = 2^{x-\frac{1}{2}} + 2^{x+\frac{1}{2}}$, then $x$ equals

Accepted Answer

Concept & Formula: To solve equations with multiple exponential terms, we need to express all terms with the same prime base. Here, the common base is 2 (since 4 = 2^2). Then, we factor out the lowest power of the base on each side to simplify the equation. a^m \cdot a^n = a^{m+n} Step-by-Step Solution: • Step 1: Convert the 4^x term to base 2. 4^x = (2^2)^x = 2^{2x} The equation becomes: 2^{2x-1} + 2^{2x} = 2^{x-(1/2)} + 2^{x+(1/2)} • Step 2: Factor out the lowest power of 2 on the left-hand side (LHS). The lowest power is 2^{2x-1}. LHS = 2^{2x-1} \cdot (1 + 2^1) LHS = 2^{2x-1} \cdot (3) • Step 3: Factor out the lowest power of 2 on the right-hand side (RHS). The lowest power is 2^{x-(1/2)}. RHS = 2^{x-(1/2)} \cdot (1 + 2^1) RHS = 2^{x-(1/2)} \cdot (3) • Step 4: Equate the simplified LHS and RHS. 3 \cdot 2^{2x-1} = 3 \cdot 2^{x-(1/2)} • Step 5: Cancel the common factor of 3 and equate the remaining exponents. 2^{2x-1} = 2^{x-(1/2)} 2x - 1 = x - (1/2) • Step 6: Solve for x. 2x - x = 1 - (1/2) x = (1/2) Exam Strategy & Shortcut: Option Verification Method: For complex exponent equations, simply plugging in the given options can be much faster than algebraic manipulation. Start with the easiest fraction, x = (1/2). LHS: 2^{2(1/2)-1} + 4^{1/2} = 2^0 + \sqrt{4} = 1 + 2 = 3 RHS: 2^{1/2 - 1/2} + 2^{1/2 + 1/2} = 2^0 + 2^1 = 1 + 2 = 3 LHS = RHS. The very first option checked is the correct one, taking less than 20 seconds. Common Pitfall: A frequent mistake is incorrectly factoring terms. For example, extracting 2^{2x} from 2^{2x-1} leaves 2^{-1} (or (1/2)), which introduces fractions early on. It's generally easier to factor out the smallest term to keep the numbers whole inside the parentheses. Final Answer: Therefore, the correct answer is (1/2).

Question 11

If $3^x = 5^y = 45^z$, then

Accepted Answer

Concept & Strategy: This is a classic "Common Constant" problem. When you have a chain of different bases equal to each other, introduce an arbitrary constant k. Express each base in terms of k, and then use the prime factorization of the largest base to link everything together into a single equation. Step-by-Step Solution: • Step 1: Set the entire equality chain equal to a constant k. 3^x = 5^y = 45^z = k • Step 2: Isolate the bases (3, 5, 45) by raising both sides to reciprocal fractional powers. 3 = k^{(1/x)} 5 = k^{(1/y)} 45 = k^{(1/z)} • Step 3: Find the multiplicative relationship between the base numbers. We know that 45 = 9 × 5 = 3^2 × 5 . • Step 4: Substitute the k-expressions into this relationship. k^{(1/z)} = (k^{(1/x)})^2 \cdot k^{(1/y)} • Step 5: Simplify the exponents using the rules (a^m)^n = a^{mn} and a^m \cdot a^n = a^{m+n}. k^{(1/z)} = k^{(2/x)} \cdot k^{(1/y)} k^{(1/z)} = k^{((2/x) + (1/y))} • Step 6: Equate the exponents, as the base k is the same. (1/z) = (2/x) + (1/y) • Step 7: Rearrange to match the given options. Move (1/y) to the other side: (2/x) = (1/z) - (1/y) Exam Strategy & Shortcut: Prime Factorization Mapping: Once you recognize the pattern A^x = B^y = C^z, look for the equation linking A, B, and C. Here, 3^2 \cdot 5^1 = 45^1. The shortcut is to directly swap the base's exponent with the reciprocal of its variable: Power of 3 is 2 \rightarrow (2/x) Power of 5 is 1 \rightarrow (1/y) Power of 45 is 1 \rightarrow (1/z) So, (2/x) + (1/y) = (1/z). Rearranging gives (2/x) = (1/z) - (1/y). You can bypass the constant k entirely on the exam. Common Pitfall: The most common trap is algebraically mismanaging the exponent rules. Students often mistakenly write k^{1/x \cdot 2} = k^{1/2x} instead of k^{2/x}. Remember that (a^b)^c = a^{bc}, which means multiplying the exponent by an integer puts the integer in the numerator. Final Answer: Therefore, the correct answer is (2/x) = (1/z) - (1/y).

Question 12

Let $r$ be the result of doubling both the base and the exponent of $a^b$, $b 
eq 0$. If $r$ equals the product of $a^b$ by $x^b$, then $x$ equals

Accepted Answer

Concept & Formula: This problem tests the fundamental rules of exponents, specifically the power of a product rule and how to equate expressions with the same exponent. The core principle is recognizing how changes to the base and exponent affect the algebraic expression. (xy)^n = x^n \cdot y^n Step-by-Step Solution: • Given: The original term is a^b. r is the result of doubling the base (from a to 2a) and the exponent (from b to 2b). Also, r = a^b \cdot x^b. • Calculation / Deduction: 1. Express r algebraically based on the first condition: r = (2a)^{2b} 2. Express the second condition mathematically: r = a^b \cdot x^b 3. Combine the two equations: (2a)^{2b} = a^b \cdot x^b 4. Simplify the RHS using the rule (xy)^n = x^n \cdot y^n: (2a)^{2b} = (ax)^b 5. To compare bases, we need the exponents to match. Rewrite the LHS so it has an exponent of b: ((2a)^2)^b = (ax)^b 6. Expand the inner term on the LHS: (4a^2)^b = (ax)^b 7. Since the exponents are identical and non-zero (b 
eq 0), we can equate the bases: 4a^2 = ax 8. Solve for x: x = (4a^2/a) = 4a Exam Strategy & Shortcut: Value Assumption (Plug-in Method): Let a = 2 and b = 1. Original term = 2^1 = 2. Double base (2 ightarrow 4) and exponent (1 ightarrow 2). So, r = 4^2 = 16. Given r = a^b \cdot x^b \implies 16 = 2^1 \cdot x^1 \implies 2x = 16 \implies x = 8. Now check the options with a = 2: (a) 2 (b) 4 (c) 2a = 2(2) = 4 (d) 4a = 4(2) = 8 Option (d) yields 8, perfectly matching our result. Common Pitfall: The most frequent error is misinterpreting "doubling both the base and the exponent". Students often write 2(a^b) or (2a)^b instead of properly squaring the entire new base and doubling the power to yield (2a)^{2b}. Final Answer: Therefore, the correct answer is 4a.

Question 13

Which of the following is the greatest?

Accepted Answer

Concept & Strategy: To compare different surds (roots) of different numbers, you cannot easily judge them in their radical form. The most reliable strategy is to convert them to fractional exponents and then raise all the terms to a common power. This common power should be the Least Common Multiple (LCM) of the denominators of the fractional exponents. x^{(1/a)} and y^{(1/b)} \rightarrow Raise to LCM(a,b) Step-by-Step Solution: • Given: The numbers are \sqrt{2}, \sqrt[3]{3}, \sqrt[4]{4}, \sqrt[6]{6}. • Calculation / Deduction: 1. Convert the radicals into fractional exponents: 2^{(1/2)}, 3^{(1/3)}, 4^{(1/4)}, 6^{(1/6)} 2. Identify the denominators of the exponents: 2, 3, 4, and 6. 3. Find the LCM of these denominators: LCM(2, 3, 4, 6) = 12 4. Raise each term to the power of 12 to eliminate the fractions, making them easy to compare: • (2^{(1/2)})^{12} = 2^{12/2} = 2^6 = 64 • (3^{(1/3)})^{12} = 3^{12/3} = 3^4 = 81 • (4^{(1/4)})^{12} = 4^{12/4} = 4^3 = 64 • (6^{(1/6)})^{12} = 6^{12/6} = 6^2 = 36 5. Compare the resulting integers: 81 > 64 = 64 > 36. 6. The greatest value is 81, which corresponds to the original term \sqrt[3]{3}. Exam Strategy & Shortcut: Quick Simplification Check: Notice immediately that \sqrt[4]{4} = (2^2)^{(1/4)} = 2^{(2/4)} = 2^{(1/2)} = \sqrt{2}. Since options (a) and (c) are mathematically identical, neither can be the unique "greatest" answer in a standard MCQ. This immediately narrows your choices down to just comparing \sqrt[3]{3} and \sqrt[6]{6}. Raise both to the power of 6: (\sqrt[3]{3})^6 = 3^2 = 9 (\sqrt[6]{6})^6 = 6^1 = 6 Since 9 > 6, \sqrt[3]{3} is clearly the greatest. Common Pitfall: Students often attempt to approximate decimal values for these roots without calculating them properly, leading to guesses. Without memorizing \sqrt[3]{3} \approx 1.442, guessing based on the base numbers (thinking 6 is the largest base so it must be greatest) is a very common trap. Final Answer: Therefore, the correct answer is \sqrt[3]{3}.

Question 14

The greatest of $\sqrt{2}, \sqrt[6]{3}, \sqrt[3]{4}, \sqrt[4]{5}$ is

Accepted Answer

Concept & Strategy: Just like other root comparison problems, the core strategy here is to eliminate the radical signs. We achieve this by converting roots into fractional exponents and then multiplying all exponents by the Least Common Multiple (LCM) of their denominators. This scales the numbers proportionally, allowing for a straightforward integer comparison. Compare a^{(1/x)}, b^{(1/y)} \implies Compare a^{\frac{LCM}{x}}, b^{\frac{LCM}{y}} Step-by-Step Solution: • Given: The values to compare are \sqrt{2}, \sqrt[6]{3}, \sqrt[3]{4}, \sqrt[4]{5}. • Calculation / Deduction: 1. Express each root as a base with a fractional exponent: 2^{(1/2)}, 3^{(1/6)}, 4^{(1/3)}, 5^{(1/4)} 2. Extract the denominators from the exponents: 2, 6, 3, and 4. 3. Calculate the LCM of (2, 6, 3, 4): LCM(2, 6, 3, 4) = 12 4. Raise every term to the power of 12 to transform them into whole numbers: • (\sqrt{2})^{12} = (2^{(1/2)})^{12} = 2^6 = 64 • (\sqrt[6]{3})^{12} = (3^{(1/6)})^{12} = 3^2 = 9 • (\sqrt[3]{4})^{12} = (4^{(1/3)})^{12} = 4^4 = 256 • (\sqrt[4]{5})^{12} = (5^{(1/4)})^{12} = 5^3 = 125 5. Compare the simplified integer values: 256 > 125 > 64 > 9. 6. The highest value is 256, which maps back to the original term \sqrt[3]{4}. Exam Strategy & Shortcut: Rapid Elimination: You can compare them in pairs to save mental load. Compare \sqrt{2} and \sqrt[3]{4}: Raise both to power 6. 2^3 = 8 vs 4^2 = 16. So \sqrt[3]{4} is greater. Compare \sqrt[6]{3} and \sqrt[4]{5}: Raise both to power 12. 3^2 = 9 vs 5^3 = 125. So \sqrt[4]{5} is greater. Now just compare the winners: \sqrt[3]{4} and \sqrt[4]{5}. Raise both to power 12. 4^4 = 256 vs 5^3 = 125. \sqrt[3]{4} is the definitive winner. Breaking it down this way avoids juggling 4 large calculations at once. Common Pitfall: A common trap is mixing up the order of the original roots when checking the options. Notice the question lists them in one order (\sqrt{2}, \sqrt[6]{3}, \sqrt[3]{4}, \sqrt[4]{5}) but the options shuffle them. Always explicitly map your final highest integer back to its exact initial radical form before selecting an option. Final Answer: Therefore, the correct answer is \sqrt[3]{4}.

Question 15

The largest number in the sequence $1, 2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 4^{\frac{1}{4}}, \ldots, n^{\frac{1}{n}}$ is

Accepted Answer

Concept & Strategy: This question touches upon a classic calculus optimization problem disguised as a sequence. The sequence represents the function f(x) = x^{(1/x)}. To find the maximum value of this sequence for integer values of x, we evaluate the function around its global maximum peak, which occurs at x = e (Euler's number, approximately 2.718). Since e falls between the integers 2 and 3, the maximum value in this integer sequence must occur at either n = 2 or n = 3. Step-by-Step Solution: • Given: The sequence n^{(1/n)} for natural numbers n = 1, 2, 3, 4 \ldots • Calculation / Deduction: 1. By mathematical theory, the sequence increases until it hits e and strictly decreases thereafter. Therefore, we only need to compare the integer values adjacent to e (which are 2 and 3). 2. Let's manually compare n=2 and n=3: We need to determine which is larger: 2^{(1/2)} or 3^{(1/3)}. 3. Take the LCM of the denominators of the powers (2 and 3), which is 6. 4. Raise both terms to the 6th power to remove fractional exponents: (2^{(1/2)})^6 = 2^3 = 8 (3^{(1/3)})^6 = 3^2 = 9 5. Since 9 > 8, we know that 3^{(1/3)} > 2^{(1/2)}. 6. Because the sequence strictly decreases for all integers n \geq 3, 3^{(1/3)} must be the absolute largest number in the entire sequence. Exam Strategy & Shortcut: Fact Memorization: For competitive exams like CAT/IIFT, the function y = x^{(1/x)} is a standard curve to memorize. Its absolute maximum always occurs at x = e \approx 2.718. Since we are dealing with a sequence of integers, the maximum must be the integer closest to 2.718, which intuitively points to 3. A quick check of 2 vs 3 confirms 3^{1/3} is indeed the peak. Common Pitfall: Many students mistakenly believe that as the base n increases, the overall value of n^{(1/n)} must also continuously increase. They might guess options (c) or (d) thinking larger numbers yield larger results. Always remember that a growing denominator in an exponent (1/n) shrinks the value drastically, overpowering the growing base. Final Answer: Therefore, the correct answer is 3^{(1/3)}.

Question 16

If $x = 5 + 2\sqrt{6}$, then $\frac{(x - 1)}{\sqrt{x}}$ is equal to

Accepted Answer

Concept & Formula: The key to simplifying expressions with surds under square roots is to express the inner surd as a perfect square. This is done by splitting the integer part into two numbers that add up to it, and whose product relates to the surd part. (a + b)^2 = a^2 + b^2 + 2ab Step-by-Step Solution: • Given: x = 5 + 2\sqrt{6} Expression to evaluate: (x - 1/\sqrt{x)} • Calculation / Deduction: 1. First, simplify the given expression by separating the terms: (x - 1/\sqrt{x)} = (x/\sqrt{x)} - (1/\sqrt{x)} = \sqrt{x} - (1/\sqrt{x)} 2. Next, find the square root of x. We need to express 5 + 2\sqrt{6} in the form of (a + b)^2. 3. Notice that 5 = 3 + 2 and \sqrt{6} = \sqrt{3 × 2}. x = 3 + 2 + 2\sqrt{3}\sqrt{2} = (\sqrt{3})^2 + (\sqrt{2})^2 + 2(\sqrt{3})(\sqrt{2}) x = (\sqrt{3} + \sqrt{2})^2 4. Taking the square root of both sides: \sqrt{x} = \sqrt{3} + \sqrt{2} 5. Now find its reciprocal, (1/\sqrt{x)}. Rationalize the denominator: (1/\sqrt{3) + \sqrt{2}} × \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \sqrt{3} - \sqrt{2} 6. Substitute these back into our simplified expression: \sqrt{x} - (1/\sqrt{x)} = (\sqrt{3} + \sqrt{2}) - (\sqrt{3} - \sqrt{2}) \sqrt{x} - (1/\sqrt{x)} = \sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2} = 2\sqrt{2} Exam Strategy & Shortcut: Algebraic Manipulation: Instead of finding \sqrt{x} first, you can manipulate the final required expression. If y = \sqrt{x} - (1/\sqrt{x)}, then squaring both sides gives y^2 = x + (1/x) - 2. Given x = 5 + 2\sqrt{6}, the reciprocal is (1/x) = 5 - 2\sqrt{6} (since 5^2 - (2\sqrt{6})^2 = 25 - 24 = 1). So, x + (1/x) = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10. Then, y^2 = 10 - 2 = 8. Thus, y = \sqrt{8} = 2\sqrt{2}. This completely avoids finding the square root of a surd! Common Pitfall: A frequent mistake is plugging the raw value of x directly into the fractional expression without simplifying it first. This leads to messy rationalization of a complex denominator like \sqrt{5 + 2\sqrt{6}}, which consumes a lot of exam time and often leads to arithmetic errors. Final Answer: Therefore, the correct answer is 2\sqrt{2}.

Question 17

Find the value of $(-2)^5 	imes (2)^{-5} 	imes (3)^3$

Accepted Answer

Concept & Formula: This problem tests the basic laws of indices, specifically dealing with negative bases raised to odd powers, and the product rule for identical bases with different exponents. (-a)^{odd} = -a^{odd} a^m × a^n = a^{m+n} Step-by-Step Solution: • Given: Expression to evaluate: (-2)^5 × (2)^{-5} × (3)^3 • Calculation / Deduction: 1. First, handle the negative sign in the first term. Since a negative number is raised to an odd power (5), the result will be negative. (-2)^5 = - (2^5) 2. Substitute this back into the original expression: - (2^5) × (2)^{-5} × (3)^3 3. Group the terms with the same base (base 2). Apply the product rule of exponents (a^m × a^n = a^{m+n}): - (2^{5 + (-5)}) × (3)^3 - (2^0) × (3)^3 4. Apply the zero exponent rule, which states that any non-zero number raised to the power of 0 is 1 (a^0 = 1): - (1) × (3)^3 5. Finally, calculate the cube of 3 (3 × 3 × 3 = 27) and multiply by -1: -1 × 27 = -27 Exam Strategy & Shortcut: Mental Math Simplification: You can solve this mentally in 5 seconds. Recognize that (-2)^5 is simply -32 and 2^{-5} is (1/32). Multiplying a number by its exact reciprocal yields 1, and preserving the negative sign gives -1. So the entire base-2 portion simply cancels out to -1. You are left with -1 × 3^3 = -27. Common Pitfall: A common error is mishandling the negative base. Students sometimes incorrectly assume that (-2)^5 × (2)^{-5} simplifies to (-2 × 2)^{5-5} or similar flawed logic, losing track of the crucial negative sign that persists purely because the exponent is an odd number. Final Answer: Therefore, the correct answer is -27.

Question 18

The exponential form of $\sqrt{\sqrt{2} 	imes \sqrt{3}}$ is

Accepted Answer

Concept & Formula: To find the exponential form of nested surds, you must apply the product rule for radicals and then convert the radical notation into fractional exponents. \sqrt{a} × \sqrt{b} = \sqrt{a × b} \sqrt[n]{x} = x^{(1/n)} Step-by-Step Solution: • Given: The expression is \sqrt{\sqrt{2} × \sqrt{3}}. • Calculation / Deduction: 1. First, simplify the innermost part of the expression by multiplying the two square roots: \sqrt{2} × \sqrt{3} = \sqrt{2 × 3} = \sqrt{6} 2. Substitute this back into the original expression: \sqrt{\sqrt{6}} 3. Convert the inner square root to a fractional exponent: \sqrt{6^{(1/2)}} 4. Convert the outer square root to a fractional exponent: (6^{(1/2)})^{(1/2)} 5. Apply the power of a power rule (x^m)^n = x^{m × n}: 6^{(1/2) × (1/2)} = 6^{(1/4)} Exam Strategy & Shortcut: Direct Transformation: You can immediately rewrite nested square roots as a 4th root. Since \sqrt{\sqrt{x}} = \sqrt[4]{x} = x^{(1/4)}, you simply multiply the inner values (2 × 3 = 6) and apply the (1/4) power. You can solve this mentally in 2 seconds. Common Pitfall: A very common mistake is adding the fractional exponents instead of multiplying them when dealing with nested roots, leading students to incorrectly choose 6^1 or 6^{(1/2)}. Always multiply powers when a power is raised to another power. Final Answer: Therefore, the correct answer is 6^{(1/4)}.

Question 19

$21^x 	imes 21^{6.5} = 21^{12.4}$

Accepted Answer

Concept & Formula: This problem tests the foundational product rule of indices. When multiplying two exponential terms with the same base, you simply add their exponents. Once the bases on both sides of an equation are identical, their exponents can be set equal to each other. a^m × a^n = a^{m+n} Step-by-Step Solution: • Given: 21^x × 21^{6.5} = 21^{12.4} • Calculation / Deduction: 1. Apply the product rule of exponents to the Left Hand Side (LHS) by combining the common base of 21: 21^{x + 6.5} = 21^{12.4} 2. Since the bases on both sides of the equation are equal (21) and non-zero, their corresponding exponents must be strictly equal: x + 6.5 = 12.4 3. Solve for the unknown variable x by subtracting 6.5 from 12.4: x = 12.4 - 6.5 x = 5.9 Exam Strategy & Shortcut: Mental Arithmetic: Recognize immediately that the base 21 is irrelevant to the actual calculation. The problem instantly simplifies in your head to x + 6.5 = 12.4. Subtraction of simple decimals (12.4 - 6.5) yields 5.9 directly. Common Pitfall: Students rushing through the exam might mistakenly divide 12.4 by 6.5, confusing the exponent rules for multiplication with the rules for power of a power. Always remember that multiplication of bases means addition of exponents. Final Answer: Therefore, the correct answer is 5.9.

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