Find the value of x.
(i) $3^2 \times (-4)^2 = (-12)^{2x}$(ii) $(\frac{9}{4})^{3} \times (\frac{8}{9})^{3}=2^{6 x}$

Given :

The given terms are,

(i) $3^2 \times (-4)^2 = (-12)^{2x}$

(ii) $(\frac{9}{4})^{3} \times (\frac{8}{9})^{3}=2^{6 x}$.

To do :

We have to find the value of x from the given terms.

Solution :

(i) $3^2 \times (-4)^2 = (-12)^{2x}$

We know that,

$a^m \times b^m = (a.b)^m$

So,  $3^2 \times (-4)^2 = (3 \times (-4))^2$

$\Rightarrow  (3 \times (-4))^2 = (-12)^{2x} $

$\Rightarrow (-12)^2 = (-12)^{2x} $

As the bases are equal, we can say,

$2 = 2x$

Rewrite,

$2x = 2$

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

Therefore, the value of x is 1.

(ii) $(\frac{9}{4})^{3} \times (\frac{8}{9})^{3}=2^{6 x}$

We know that,

$a^m \times b^m = (a.b)^m$

So, $(\frac{9}{4})^{3} \times (\frac{8}{9})^{3}=(\frac{9}{4} \times \frac{8}{9})^{3}$

$\Rightarrow (\frac{9}{4} \times \frac{8}{9})^{3} = 2^{6 x}$

$\Rightarrow (2)^3 = 2^{6x}$

As the bases are equal, we can say,

$3 = 6x$

Rewrite,

$6x = 3$

$x = \frac{3}{6}$

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

Therefore, the value of x is $\frac{1}{2}$.

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Updated on: 10-Oct-2022

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