Question

If $4^x+2^{2x-1}=3^{x+\frac{1}{2}}+3^{x-\frac{1}{2}}$, then x is equal to

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $\frac{5}{2}$
• D. $1$

Answer 1

The correct option is B $\frac{3}{2}$

Given that,

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

$4^x+\frac{4^x}{2}=3^x\left(3^{\frac{1}{2}}+\frac{1}{3^{\frac{1}{2}}}\right)$

$4^x\left(1+\frac{1}{2}\right)=3^x\left(\frac{3+1}{3^{\frac{1}{2}}}\right)$

$4^x\left(\frac{3}{2}\right)=3^x\left(\frac{4}{3^{\frac{1}{2}}}\right)$

$\frac{4^x}{3^x}=\left(\frac{4}{3^{\frac{1}{2}}}\right)\left(\frac{2}{3}\right)$

${\left[\frac{4}{3}\right]}^x=\left(\frac{4^{\frac{3}{2}}}{3^{\frac{3}{2}}}\right)$

$x=\frac{3}{2}$