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If $9^{2x-1}-81^{x-1}= 1944$ then $x$ is

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

Given that, $9^{2x-1} – 81^{x-1} = 1944$

$\Rightarrow 9^{2x-1} – 9^{2x-2} = 1944$

$\Rightarrow 9^{2x}\cdot 9^{-1} – 9^{2x}\cdot9^{-2} = 1944$

$\Rightarrow 9^{2x}\left(\frac{1}{9}-\frac{1}{81}\right) = 1944$

$\Rightarrow 9^{2x}\left(\frac{8}{81}\right) = 1944$

$\Rightarrow 9^{2x} = \frac{1944}{8} \cdot 81$

$\Rightarrow (3^{2})^{2x} = 243 \cdot 81$

$\Rightarrow 3^{4x} = 3^{5} \cdot 3^{4}$

$\Rightarrow 3^{4x} = 3^{9}$

The base is same on both sides, then equating the powers.

$\Rightarrow 4x = 9$

$\Rightarrow \boxed{ x = \frac{9}{4}}.$

Correct Answer $:\text{B}$
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