Question

Find x, if : ${\displaystyle \sqrt{2^{x+3}}=16}$

Answer 1

${\displaystyle \sqrt{2^{x+3}}=16}$

${\displaystyle {\left(2^{x+3}\right)}^{\frac{1}{2}}=2\times 2\times 2\times 2}$

⇒ ${\displaystyle {\left(2\right)}^{\frac{x+3}{2}}=2^4}$

We know that if bases are equal, the powers are equal.
⇒ ${\displaystyle \frac{x+3}{2}=4}$

⇒ x + 3 = 8
⇒ x = 8 - 3
⇒ x = 5