Question

If $\frac{4^{\left(x^2\right)}}{{16}^x}=1$. Find the value of $x$ where $x\ne 0$.

• A. 2
• B. 3
• C. 1
• D. any of the above

Answer 1

The correct option is A

2

Given, $\frac{4^{\left(x^2\right)}}{{16}^x}=1$

$\Rightarrow 4^{x^2}={16}^x$

$\Rightarrow 4^{x^2}=(4^2{)}^x$

$\Rightarrow 4^{x^2}=4^{2x}$

Since bases are equal powers should also be equal

$\Rightarrow \left(x^2\right)=2x$

$\Rightarrow x=2$