Question

Não consigo resolver, alguém me ajuda?
2^x-3+2^x-1+2^x=52

Answer 1

Boa tarde Nicole!

Solução!

$2^{x-3}+2^{x-1} +2^{x}=52\\\\\\ 2^{x}.2^{-3}+2^{x}.2^{-1}+ 2^{x}=52\\\\\\\ 2^{x}(2^{-3}+2^{-1}+1)=52\\\\\\\ 2^{x}\bigg(\bigg( \dfrac{1}{2}\bigg)^{3}+\bigg( \dfrac{1}{2}\bigg)^{1}+1\bigg)=52\\\\\\\ 2^{x}\bigg( \dfrac{1}{8}+\dfrac{1}{2}+1\bigg)=52\\\\\\\ 2^{x}\bigg( \dfrac{1+4+8}{8}\bigg)=52\\\\\\\\\ 2^{x}\bigg( \dfrac{13}{8}\bigg)=52$


$2 ^{x}= \dfrac{52}{ \dfrac{13}{8} }\\\\\\\\ 2 ^{x}= 52\times \dfrac{8}{13}\\\\\\\ 2^{x}= \dfrac{416}{13}\\\\\\\ 2^{x}=32\\\\\\ 2^{x}=2^{5}\\\\\ Bases~~iguais!\\\\\\\ x=5$


$\boxed{Resposta:~~2^{x-3}+2^{x-1} +2^{x}=52~~\Rightarrow~~S=\{5\}}$

Boa tarde!
Bons estudos!

Answer 2

Ae mano,

use a propriedade da exponenciação:

$\large\boxed{a^{m+n}=a^m\cdot a^n}$

Faça o mesmo com os termos desta equação exponencial:

$2^{x-3}+2^{x-1}+2^x=52\\ 2^x\cdot2^{-3}+2^x\cdot2^{-1}+2^x=52\\\\ 2^x\cdot \dfrac{1}{2^3}+2^x\cdot \dfrac{1}{2^1}+2^x=52\\\\ \dfrac{1}{8}\cdot2^x+ \dfrac{1}{2}\cdot2^x+2^x=52$

Agora, mude a variável $\large\boxed{\underbrace{\overbrace{2^x=y}}}$

$\dfrac{1}{8}\cdot2^x+ \dfrac{1}{2}\cdot2^x+2^x=32\\\\ \dfrac{1}{8}y+ \dfrac{1}{2}y+y=52\\\\ \dfrac{1}{8}y+ \dfrac{1}{2}y+ \dfrac{2}{2}y= \dfrac{104}{2} \\\\ m.m.c.~de~~ 2~e~8=8\\\\ \dfrac{1\cdot y+4\cdot y+4\cdot2y}{\not8}= \dfrac{4\cdot104}{\not8}\\\\ y+4y+8y=416\\ 13y=416\\\\ y= \dfrac{416}{16}\\\\ y=32$

Descobrimos que y vale 32, vamos retomar a variável
 
original, $\large\boxed{\underbrace{\overbrace{2^x=y}}}$

$2^x=32\\ 2^x=2^5\\ \not2^x=\not2^5\\\\ x=5\\\\\\ \huge\boxed{\text{S}=\{5\}}$

Tenha ótimos estudos ;D