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Solução!
![log_{16} \frac{1}{4} \\\\\\
log_{16} \frac{1}{4}=x \\\\\\
16^{x}= \frac{1}{4} \\\\\\\
(2^{4})^{x} = \frac{1}{2^{2} }\\\\\\\
(2)^{4x}=2^{-2}\\\\\\
4x=-2\\\\\\\
x= \dfrac{-2}{4} \\\\\\\\
\boxed{x=- \dfrac{1}{2}}](https://tex.z-dn.net/?f=log_%7B16%7D+%5Cfrac%7B1%7D%7B4%7D+%5C%5C%5C%5C%5C%5C%0A+log_%7B16%7D+%5Cfrac%7B1%7D%7B4%7D%3Dx+%5C%5C%5C%5C%5C%5C%0A16%5E%7Bx%7D%3D+%5Cfrac%7B1%7D%7B4%7D+%5C%5C%5C%5C%5C%5C%5C%0A%282%5E%7B4%7D%29%5E%7Bx%7D+%3D+%5Cfrac%7B1%7D%7B2%5E%7B2%7D+%7D%5C%5C%5C%5C%5C%5C%5C%0A%282%29%5E%7B4x%7D%3D2%5E%7B-2%7D%5C%5C%5C%5C%5C%5C%0A4x%3D-2%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B-2%7D%7B4%7D+%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bx%3D-+%5Cdfrac%7B1%7D%7B2%7D%7D++++++ "log_{16} \frac{1}{4} \\\\\\
log_{16} \frac{1}{4}=x \\\\\\
16^{x}= \frac{1}{4} \\\\\\\
(2^{4})^{x} = \frac{1}{2^{2} }\\\\\\\
(2)^{4x}=2^{-2}\\\\\\
4x=-2\\\\\\\
x= \dfrac{-2}{4} \\\\\\\\
\boxed{x=- \dfrac{1}{2}}")
Boa tarde!
Bons estudos!
Solução!
Boa tarde!
Bons estudos!
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