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Is there an algebraic trick using some change of variable or properties of exponent or logarithms to solve the following equation algebraically for $x\in \mathbb{R}$?

$\log_{\frac{1}{4}}(x)=(\frac{1}{4})^x$

user13892
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1 Answers1

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Put $y = 4^{-x}$, then you want to solve $4^{-y} = y$ (and it follows that $y=x$). A unique solution exists because $4^{-y}-y$ has negative derivative and is positive at $0$, negative at $1$ (so the solution is between $0, 1$). One can see that $y=1/2$ is a solution, so $x = 1/2$.

red whisker
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