Question

Função Exponencial

$\sqrt{4} ^{x+1} = \sqrt[3]{16}$

Answer 1

$\sqrt{ 4^{x+1} } = \sqrt[3]{16}$

$\sqrt{ ( 2^{2} )^{x+1} } = \sqrt[3]{ 2^{4} }$

$\sqrt{ ( 2 )^{2.(x+1)} } = \sqrt[3]{ 2^{4} }$

$2^\frac{2.(x+1)}{2}= 2^{ \frac{4}{3} }$

$\dfrac{2.(x+1)}{2}=\dfrac{4}{3}$

$x+1=\dfrac{4}{3}$

$x=\dfrac{4}{3}-1$

$x=\dfrac{4-3}{3}$

$x=\dfrac{1}{3}$