Question

If $4^{x+3}\times 2^{x-3}-128=0$ then the value of $x$ is :

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{3}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is D $\frac{4}{3}$

We have,

$4^{x+3}\times 2^{x-3}-128=0$

$\Rightarrow {\left(2\right)}^{2(x+3)}\times 2^{x-3}-128=0$

$\Rightarrow {\left(2\right)}^{2x+6+x-3}-128=0$

$\Rightarrow {\left(2\right)}^{3x+3}-128=0$

$\Rightarrow {\left(2\right)}^{3x+3}=128$

$\Rightarrow {\left(2\right)}^{3x+3}={\left(2\right)}^7$

On comparing both side and we get,

$3x+3=7$

$\Rightarrow 3x=7-3$

$\Rightarrow 3x=4$

$\Rightarrow x=\frac{4}{3}$

Hence, this is the answer.