Skip to main content

Solve for x,

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

We have, ${2^{2x}} + \frac{{{2^{2x}}}}{2} = {3^x}\sqrt 3  + \frac{{{3^x}}}{{\sqrt 3 }}$

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

$ \Rightarrow \frac{{{2^{2x}}}}{8} = \frac{{{3^x}}}{{3\sqrt 3 }}$

$ \Rightarrow {2^{2x - 3}} = {3^{x - \frac{3}{2}}}$

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

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

The above is only possible if $x - \frac{3}{2} = 0$ or $x=\frac {3}{2}$.