s=log1/7 na base 49+log1/125na base 125-log5/3 na base 0,6
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Ae mano,
use a definição de logaritmos:
![\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]](https://tex.z-dn.net/?f=%5Ctext%7BS%7D%3D%5Clog_%7B49%7D%5Cleft%28+%5Cdfrac%7B1%7D%7B7%7D+%5Cright%29%2B%5Clog_%7B125%7D%5Cleft%28+%5Cdfrac%7B1%7D%7B125%7D%5Cright%29-%5Clog_%7B0%2C6%7D%5Cleft%28+%5Cdfrac%7B5%7D%7B3%7D%5Cright%29%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D%5Cleft%5B49%5Ex%3D+%5Cdfrac%7B1%7D%7B7%7D%5Cright%5D%2B%5Cleft%5B125%5Ex%3D+%5Cdfrac%7B1%7D%7B125%7D%5Cright%5D-%5Cleft%5B0%2C6%5Ex%3D+%5Cdfrac%7B5%7D%7B3%7D%5Cright%5D%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D%5Cleft%5B%287%5E2%29%5Ex%3D+%5Cdfrac%7B1%7D%7B7%5E1%7D+%5Cright%5D%2B%5Cleft%5B%285%5E3%29%5Ex%3D+%5Cdfrac%7B1%7D%7B5%5E3%7D%5Cright%5D-%5Cleft%5B%5Cleft%28+%5Cdfrac%7B6%7D%7B10%7D%5Cright%29%5Ex%3D+%5Cdfrac%7B5%7D%7B3%7D%5Cright%5D%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D%5B7%5E%7B2x%7D%3D7%5E%7B-1%7D%5D%2B%5B5%5E%7B3x%7D%3D5%5E%7B-3%7D%5D-%5Cleft%5B%5Cleft%28+%5Cdfrac%7B6%5Cdiv2%7D%7B10%5Cdiv2%7D+%5Cright%29%5Ex%3D+%5Cdfrac%7B5%7D%7B3%7D+%5Cright%5D "\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}}}](https://tex.z-dn.net/?f=%5Ctext%7BS%7D%3D%5B%5Cnot7%5E%7B2x%7D%3D%5Cnot7%5E%7B-1%7D%5D%2B%5B%5Cnot5%5E%7B3x%7D%3D%5Cnot5%5E%7B-3%7D%5D-%5Cleft%5B%5Cleft%28+%5Cdfrac%7B3%7D%7B5%7D+%5Cright%29%5Ex%3D+%5Cdfrac%7B5%7D%7B3%7D%5E1+%5Cright%5D%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D%5B2x%3D-1%5D%2B%5B3x%3D-3%5D-%5Cleft%5B%5Cleft%28+%5Cdfrac%7B%5Cnot5%7D%7B%5Cnot3%7D%5Cright%29%5E%7B-x%7D%3D+%5Cdfrac%7B%5Cnot5%7D%7B%5Cnot3%7D%5E1%5Cright%5D%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D%5Cleft%5Bx%3D+%5Cdfrac%7B1%7D%7B2%7D%5Cright%5D%2B%5Cleft%5Bx%3D+%5Cdfrac%7B%7E%7E3%7D%7B-3%7D%5Cright%5D-%5B-x%3D1%5D%5C%5C%5C%5C%0A%5Ctext%7BS%7D%3D+%5Cdfrac%7B1%7D%7B2%7D-1%2B1%5C%5C%5C%5C%5C%5C%0A%5Chuge%5Cboxed%7B%5Cboxed%7B%5Ctext%7BS%7D%3D+%5Cdfrac%7B1%7D%7B2%7D%7D%7D++++ "\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
use a definição de logaritmos:
TENHA ÓTIMOS ESTUDOS ;D
Lucianofurman:
valeu cara
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