Question

Solve: $$ 4^x - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1} $$

• A. \( \frac{5}{2} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{2} \)
• D. \( \frac{7}{2} \)

Hint:

Try suitable substitutions when both sides have exponential expressions with matching bases. Rational exponents often simplify nicely.

Answer 1

The Correct Option is C

Simplify and rewrite terms using exponents: \[ 4^x = (2^2)^x = 2^{2x}, \quad 2^{2x - 1} = \frac{2^{2x}}{2} \] \[ 3^{x - \frac{1}{2}} = \frac{3^x}{\sqrt{3}}, \quad 3^{x + \frac{1}{2}} = 3^x \sqrt{3} \] Substitute into the equation: \[ 2^{2x} - \frac{3^x}{\sqrt{3}} = 3^x \sqrt{3} - \frac{2^{2x}}{2} \] Move all terms to one side: \[ 2^{2x} + \frac{2^{2x}}{2} = 3^x \sqrt{3} + \frac{3^x}{\sqrt{3}} \Rightarrow \frac{3}{2} \cdot 2^{2x} = 3^x \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) \] Simplify RHS: \[ \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \] So: \[ \frac{3}{2} \cdot 2^{2x} = 3^x \cdot \frac{4}{\sqrt{3}} \] Try \( x = \frac{3}{2} \): LHS: \[ \frac{3}{2} \cdot 2^{3} = \frac{3}{2} \cdot 8 = 12 \] RHS: \[ 3^{3/2} \cdot \frac{4}{\sqrt{3}} = (\sqrt{3})^3 \cdot \frac{4}{\sqrt{3}} = 3\sqrt{3} \cdot \frac{4}{\sqrt{3}} = 12 \] Both sides equal ⇒ solution is: \[ \boxed{x = \frac{3}{2}} \]