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

Question

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

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

Answer 1

Yes because $b^{\log_{b} y} = b^{\log_{b} x} \Longleftrightarrow y = x$.

Answer 2

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.

Answer 3

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

Answer 4

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