Question

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

Answer 1

We have,

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

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

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

$\Rightarrow {2.4}^x=3^x[\sqrt{3}+\frac{1}{\sqrt{3}}]$

$\Rightarrow {2.4}^x=3^x\left[\frac{3+1}{\sqrt{3}}\right]$

$\Rightarrow {2.4}^x=\frac{3^x}{\sqrt{3}}.4$

$\Rightarrow 4^x=\frac{3^x}{\sqrt{3}}.2$

$\Rightarrow \frac{4^x}{3^x}=\frac{2}{\sqrt{3}}$

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

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

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

Comparing power and we get,

$2x=1$

$x=\frac{1}{2}$

Hence, this is the answer.