Question

Resolva, em IR, a equação:
2x – 1 + 2x + 2x + 1 – 2x + 2 + 2x + 3 = 120

Answer 1

Olá,

dada a equação exponencial,

$\largege\boxed{2^{x-1}+2^x+2^{x+1}-2^{x+2}+2^{x+3}=120}$

podemos desmembrar as potências de base 2, e usarmos a propriedade da exponenciação:

$a^{m+n}=a^m\cdot a^n\\\\2^x\cdot2^{-1}+2^x+2^x\cdot2^1-2^x\cdot2^2+2^x\cdot2^3=120\\\\pomos~2^x~em~evidencia:\\\\ 2^x\cdot(2^{-1}+1+2^1-2^2+2^3)=120\\\\ 2^x\cdot\left( \dfrac{1}{2}+1+2-4+8\right)=120\\\\ 2^x\cdot \dfrac{15}{2}=120\\\\ 2^x= \dfrac{120}{1}:\dfrac{15}{2}\\\\ 2^x= \dfrac{120}{1}\cdot \dfrac{2}{15}\\\\ 2^x= \dfrac{240}{15}\\\\ 2^x=16\\ 2^x=2^4\\ \not2^x=\not2^4\\ x=4\\\\\\ \Large\boxed{\boxed{S=\{4\}}}$

Tenha ótimos estudos, e qualquer dúvida me chame ;D

Answer 2

Resposta:

[t$[tex]a^{m+n}=a^m\cdot a^n\\\\2^x\cdot2^{-1}+2^x+2^x\cdot2^1-2^x\cdot2^2+2^x\cdot2^3=120\\\\pomos~2^x~em~evidencia:\\\\ 2^x\cdot(2^{-1}+1+2^1-2^2+2^3)=120\\\\ 2^x\cdot\left( \dfrac{1}{2}+1+2-4+8\right)=120\\\\ 2^x\cdot \dfrac{15}{2}=120\\\\ 2^x= \dfrac{120}{1}:\dfrac{15}{2}\\\\ 2^x= \dfrac{120}{1}\cdot \dfrac{2}{15}\\\\ 2^x= \dfrac{240}{15}\\\\ 2^x=16\\ 2^x=2^4\\ \not2^x=\not2^4\\ x=4\\\\\\ \Large\boxed{\boxed{S=\{4\}}}$[/tex]

Explicação passo-a-passo: