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

### 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: $$ 2^{2x-1} + 2^{2x} = 2^{x-\frac{1}{2}} + 2^{x+\frac{1}{2}} $$ * **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 $2^{x-\frac{1}{2}}$. RHS $= 2^{x-\frac{1}{2}} \cdot (1 + 2^1) $ RHS $= 2^{x-\frac{1}{2}} \cdot (3) $ * **Step 4:** Equate the simplified LHS and RHS. $$ 3 \cdot 2^{2x-1} = 3 \cdot 2^{x-\frac{1}{2}} $$ * **Step 5:** Cancel the common factor of $3$ and equate the remaining exponents. $$ 2^{2x-1} = 2^{x-\frac{1}{2}} $$ $$ 2x - 1 = x - \frac{1}{2} $$ * **Step 6:** Solve for $x$. $$ 2x - x = 1 - \frac{1}{2} $$ $$ x = \frac{1}{2} $$ ### 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, $x = \frac{1}{2}$. 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 $\frac{1}{2}$), 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 **$\frac{1}{2}$**.