Question

Resolva em IR a equação exponencial

$\LARGE\boxed{\boxed{\boxed{2 ^{ \sqrt{x+7} }+3\cdot4^x-56=0}}}|\\-$

Ó aí pra vcs!!!

Answer 1

Temos que, $2^{\sqrt{x+7}}+3\cdot4^{x}-56=0$.

Assim, $2^{\sqrt{x+7}}+3\cdot4^{x}=56$.

Observemos que, $56=2^3+3\cdot4^2$.

Deste modo, $2^{\sqrt{x+7}}=2^{3}$ e obtemos:

$\sqrt{x+7}=3~~\Rightarrow~~x+7=9~~\Rightarrow~~x=2$

Analogamente, $3\cdot4^{2}=3\cdot(x^2)~~\Rightarrow~~x=2$.

$S=\{2\}$.

Answer 2

$2 ^{ \sqrt{x+7} }+3.4^x-56=0 \\ 2 ^{ \sqrt{x+7} }+3.4^x = 56$

$2 ^{ \sqrt{x+7} }+3.4^x = {2}^{3} + 3. {4}^{2}$

$\sqrt{x + 7} = 3 \\ { (\sqrt{x + 7})}^{2} = {3}^{2} \\ x + 7 = 9 \\ x = 9 - 7 = 2$

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

$\boxed{\boxed{\mathsf{s=\{2\}}}}$