Question

s=log1/7 na base 49+log1/125na base 125-log5/3 na base 0,6

Answer 1

Ae mano,

use a definição de logaritmos:

$\Large\boxed{\log_a(b)=c\Rightarrow a^c=b}$

$\text{S}=\log_{49}\left( \dfrac{1}{7} \right)+\log_{125}\left( \dfrac{1}{125}\right)-\log_{0,6}\left( \dfrac{5}{3}\right)\\\\ \text{S}=\left[49^x= \dfrac{1}{7}\right]+\left[125^x= \dfrac{1}{125}\right]-\left[0,6^x= \dfrac{5}{3}\right]\\\\ \text{S}=\left[(7^2)^x= \dfrac{1}{7^1} \right]+\left[(5^3)^x= \dfrac{1}{5^3}\right]-\left[\left( \dfrac{6}{10}\right)^x= \dfrac{5}{3}\right]\\\\ \text{S}=[7^{2x}=7^{-1}]+[5^{3x}=5^{-3}]-\left[\left( \dfrac{6\div2}{10\div2} \right)^x= \dfrac{5}{3} \right]$

$\text{S}=[\not7^{2x}=\not7^{-1}]+[\not5^{3x}=\not5^{-3}]-\left[\left( \dfrac{3}{5} \right)^x= \dfrac{5}{3}^1 \right]\\\\ \text{S}=[2x=-1]+[3x=-3]-\left[\left( \dfrac{\not5}{\not3}\right)^{-x}= \dfrac{\not5}{\not3}^1\right]\\\\ \text{S}=\left[x= \dfrac{1}{2}\right]+\left[x= \dfrac{~~3}{-3}\right]-[-x=1]\\\\ \text{S}= \dfrac{1}{2}-1+1\\\\\\ \huge\boxed{\boxed{\text{S}= \dfrac{1}{2}}}$

TENHA ÓTIMOS ESTUDOS ;D