Question

10^logx+2^0=log(10/2)

Answer 1

Olá,

use as propriedades:

$\log_b(a)^n\Rightarrow n\cdot \log_b(a)\\\\ \log_b\left( \dfrac{a}{c}\right)\Rightarrow\log_b(a)-\log_b(c)\\\\ \log_b(a)+\log_b(c)\Rightarrow \log_b(a\cdot c)\\\\ \log_b(b)=1$

$10^{\log(x+2^0)}=\log\left( \dfrac{10}{2}\right)\\ \log_{10}[10^{\log(x+1)}]=\log_{10}\left( \dfrac{10}{2}\right)\\ \log_{10}(10)^{\log_{10}(x+1)}=\log_{10}(10)-\log_{10}2\\ \log_{10}(x+1)\cdot\log_{10}10=1-\log_{10}2\\ 1\cdot\log_{10}(x+1)=1-\log_{10}2\\ \log_{10}(x+1)=1-\log_{10}2\\ \log_{10}(x+1)+\log_{10}2=1\\ \log_{10}[(x+1)\cdot2]=1\\ \log_{10}(2x+2)=1\\\\ 2x+2=10^1\\ 2x+2=10\\ 2x=10-2\\ 2x=8\\\\ x= \dfrac{8}{2} \\\\ x=4\\\\\\ \huge\boxed{S=\{4\}}$

Tenha ótimos estudos ;D