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I hit a snag whilst revising some log rules, could anyone confirm my suspicion:

$$\log _b \left( x \right) = \log _b \left( y \right) \rightarrow x = y ?$$

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Yes, $\log^{-1} = \exp$, so $\log$ is necessarily injective. –  Alexei Averchenko Feb 25 '11 at 16:31
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Yes because $b^{\log_{b} y} = b^{\log_{b} x} \Longleftrightarrow y = x$.

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But why? ${ }{}{}{}$ –  Pedro Tamaroff Jun 20 '12 at 1:54
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Yes it does, by the following argument: Suppose that $\mathrm{log}_b(x) = \mathrm{log}_b(y)$. Then $b^{\mathrm{log}_b(x)} = b^{\mathrm{log}_b(y)}$. But now (by the definition of $\mathrm{log_b(\cdot)}$) we know that $b^{\mathrm{log}_b(x)} = x$, so we conclude that $x = y$ as required.

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By definition, $b^{\log_b t} = t$.

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$\log(x)$ is increasing, for instance its derivative $1/x$ is positive.

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$\log_b(x)$ is not increasing for $0 \lt b \lt 1$ though it is still monotonic –  Henry Mar 14 '12 at 0:18
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