Question

The value of $u$ which satisfies $\frac{{16}^u}{2^{2u}}=4^2$ is .

Answer 1

$\underset{–}{\text{Need to find:}}$ Value of $u$ which satisfies the exponential equation $\frac{{16}^u}{2^{2u}}=4^2$

$\frac{{16}^u}{2^{2u}}=4^2$
$\Rightarrow \frac{(2^4{)}^u}{2^{2u}}=(2^2{)}^2$
$[\because 16=2\times 2\times 2\times 2=2^4$ and $4=2\times 2=2^2$ ]

Applying exponent rule of multiplication: $(a^m{)}^n=a^{m\times n}$

$\Rightarrow \frac{2^{4\times u}}{2^{2u}}=2^{2\times 2}$
$\Rightarrow \frac{2^{4u}}{2^{2u}}=2^4$

Applying exponent rule of division: $\frac{a^m}{a^n}=a^{m-n}$

$\Rightarrow 2^{4u-2u}=2^4$
$\Rightarrow 2^{2u}=2^4$

In the above exponential equation, the base $\left(2\right)$ is same. Hence, for L.H.S=R.H.S, the powers must be same.

$\Rightarrow 2u=4$
$\Rightarrow \frac{2u}{2}=\frac{4}{2}$
$\Rightarrow u=\frac{2\times 2}{2}$
$\Rightarrow \underset{–}{u=2}$

$\therefore$ The value of $u$ which satisfies the exponential equation $\frac{{16}^u}{2^{2u}}=4^2$ is $\underset{–}{2}$.