Question

Resolva a equação: 2^(1+x) + √8 = √72

consigo desenvolver até certa parte, mas não até a resposta: 3/2

Answer 1

$2^{1+x}+\sqrt{8}=\sqrt{72}\\\\2^{x+1}=\sqrt{72}-\sqrt{8}\\\\2^{x+1}=\sqrt{36\cdot2}-\sqrt{4\cdot2}\\\\2^{x+1}=\sqrt{36}\sqrt{2}-\sqrt{4}\sqrt{2}\\\\2^{x+1}=6\sqrt{2}-2\sqrt{2}\\\\2^{x+1}=4\sqrt{2}\\\\2^{x+1}=2^{2}\cdot2^{1/2}\\\\2^{x+1}=2^{2+(1/2)}$

Bases iguais, iguale os expoentes:

$x+1=2+\dfrac{1}{2}\\\\\\x=2+\dfrac{1}{2}-1\\\\\\x=1+\dfrac{1}{2}\\\\\\x=\dfrac{2+1}{2}\\\\\\\boxed{\boxed{x=\dfrac{3}{2}}}$

Answer 2

2^( 1 + x) = 6V2 - 2V2
2^( 1 + x) = 4 $\sqrt{2}$
2^( 1 + x) = 2^2$\sqrt{2}$
2^( 1 + x) = 2^2.2^1/2
2^( 1 + x) = 2^5/2
1+x = 5/2
x= 5/2 - 1
x= 3/2