Question 1

If $\log 3, \log (3^x - 2)$ and $\log (3^x + 4)$ are in arithmetic progression, then $x$ is equal to

Accepted Answer

Concept & Formula: This problem combines Logarithmic properties with the rule of Arithmetic Progressions (AP). If three terms A, B, C are in an arithmetic progression, the middle term is the average of the other two, which gives the fundamental AP formula: 2B = A + C We will also use the logarithmic power and product rules: 1. n \log x = \log(x^n) 2. \log x + \log y = \log(xy) Step-by-Step Solution: Given the terms are in AP: 2 \log(3^x - 2) = \log 3 + \log(3^x + 4) Step 1: Apply the log rules to consolidate the equation. Bring the 2 into the exponent on the left, and multiply the arguments on the right. \log((3^x - 2)^2) = \log(3(3^x + 4)) Step 2: Since the logs are equal, their arguments must be equal. Drop the logarithms. (3^x - 2)^2 = 3(3^x + 4) Step 3: Substitute y = 3^x to make it a standard quadratic equation. (y - 2)^2 = 3(y + 4) Step 4: Expand both sides. y^2 - 4y + 4 = 3y + 12 Step 5: Move all terms to one side to solve the quadratic. y^2 - 7y - 8 = 0 Step 6: Factor the equation. (y - 8)(y + 1) = 0 So, y = 8 or y = -1. Step 7: Substitute 3^x back in for y. Since an exponential function (3^x) is always positive for real numbers, 3^x = -1 is impossible and must be rejected. 3^x = 8 Step 8: Solve for x by converting to logarithmic form. x = \log_{3} 8 Exam Strategy & Shortcut: Option Verification: Instead of solving the quadratic, you can test the options directly into 3^x. If you test option (b) where x = \log_{3} 8, then 3^x simply becomes 8. The original sequence becomes: \log 3, \log(8-2), \log(8+4), which simplifies to \log 3, \log 6, \log 12. Check the common difference: \log 6 - \log 3 = \log(6/3) = \log 2 \log 12 - \log 6 = \log(12/6) = \log 2 The common differences match, instantly confirming option (b) is correct. Common Pitfall: The main trap is forgetting to reject the negative root (y = -1). If students don't remember that a^x > 0 for positive bases, they might get confused or look for an option that accommodates both roots. Always discard negative exponential outputs. Final Answer: Therefore, the correct answer is \log_{3} 8.

Question 2

If $\log_{10} a = p$ and $\log_{10} b = q$, then what is $\log_{10} (a^p b^q)$ equal to?

Accepted Answer

Concept & Formula: This question tests the ability to fully expand a logarithmic expression using fundamental properties and substitute given algebraic values. Formulas used: 1. Product Rule: \log(xy) = \log x + \log y 2. Power Rule: \log(x^n) = n \log x Step-by-Step Solution: Given the expression: \log_{10} (a^p b^q) Step 1: Apply the logarithmic Product Rule to separate the variables being multiplied inside the argument. \log_{10} (a^p) + \log_{10} (b^q) Step 2: Apply the logarithmic Power Rule to bring the exponents (p and q) to the front of their respective terms as coefficients. p \log_{10} a + q \log_{10} b Step 3: Substitute the known values provided in the question (\log_{10} a = p and \log_{10} b = q). p(p) + q(q) Step 4: Multiply the terms together to get the final algebraic expression. p^2 + q^2 Exam Strategy & Shortcut: Direct Visual Mapping: This is a pure property-based question that should require zero calculation time. Read the exponents as coefficients immediately in your head: a^p gives p × \log a (which maps to p × p). b^q gives q × \log b (which maps to q × q). Because it is a product (a^p × b^q), you add the two results. The answer p^2 + q^2 should be visually apparent without writing down a single intermediate step. Common Pitfall: A severe mistake is misinterpreting the product rule and assuming that \log(a^p b^q) equals (\log a^p) × (\log b^q). This incorrect multiplication of terms leads directly to the trap option (c) p^2q^2. Remember: The log of a product is the *sum* of the individual logs. Final Answer: Therefore, the correct answer is p^2 + q^2.

Logarithm Questions

Questions

Practice smarter