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

### Sum of the coefficients in the expansion of $(x+y)^n$ ....

If the sum of the coefficients in the expansion of $(x+y)^n$ is 4096, then the greatest coefficient in the expansion is _ _ _ _ . Solution $C_0 + C_1 + C_2 + C_3 + ......................... + C_n =4096$ $\therefore 2^n = 4096 =2^{12}$ $\Rightarrow n = 12$ Greatest coefficient = ${}^{12}{C_6} = 924$