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If $9^{\left ( x-1/2 \right )}-2^{\left ( 2x-2 \right )}=4^{x}-3^{\left (2x-3 \right )}$, then $x$ is

1. $3/2$
2. $2/5$
3. $3/4$
4. $4/9$

### 1 comment

Given that,  $9^{\left(x−\frac{1}{2}\right)}−2^{(2x−2)}=4^{x}−3^{(2x−3)}$

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

$\Rightarrow 3^{(2x-1)}-2^{(2x−2)}=2^{2x}−3^{(2x−3)}$

$\Rightarrow (3^{2x-1})+3^{(2x−3)}=2^{2x}+2^{(2x−2)}$

$\Rightarrow 3^{2x}\cdot3^{−1}+3^{2x}\cdot3^{−3}=2^{2x}+2^{2x}\cdot2^{−2}$

$\Rightarrow \dfrac{3^{2x}}{3}+\dfrac{3^{2x}}{27}=2^{2x}+\dfrac{2^{2x}}{4}$

$\Rightarrow \dfrac{9 \cdot 3^{2x}+3^{2x}}{27}=\dfrac{4 \cdot2^{2x}+2^{2x}}{4}$

$\Rightarrow \dfrac{3^{2x}(9+1)}{27}=\dfrac{2^{2x}(4+1)}{4}$

$\Rightarrow \dfrac{10}{27}(3^{2x})=\dfrac{5}{4}(2^{2x})$

$\Rightarrow \dfrac{3^{2x}}{27}=\dfrac{2^{2x}}{8}$

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

$\Rightarrow \boxed{ 3^{(2x-3)}=2^{(2x-3)}}$

This is only possible when, $2x-3=0$

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

Correct Answer $:\text{A}$
11.2k points
$\implies 9^\left ( x-1/2 \right )-2^{2x-2}=4^x-3^{2x-3}$

$\implies 3^{2\left(x-½ \right)}-2^{2x-2}=2^{2x}-3^{2x-3}$

$\implies 3^{2x-1}-2^{2x-2}=2^{2x}-3^{2x-3}$

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

$\implies \frac{3^{2x}}{3}+\frac{3^{2x}}{27}=2^{2x}+\frac{2^{2x}}{4}$

$\implies 3^{2x}(\frac{1}{3}+\frac{1}{27})=2^{2x}(1+\frac{1}{4})$

$\implies 3^{2x}*\frac{10}{27}=2^{2x}*\frac{5}{4}$

$\implies \frac{3^{2x}}{2^{2x}}=\frac{27}{8}$

$\implies (\frac{3}{2})^{2x}=(\frac{3}{2})^3$

$\implies 2x=3$

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

option (A) is correct.
6.0k points

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