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:

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

Yes, $\log^{-1} = \exp$, so $\log$ is necessarily injective. — Aleksei Averchenko, Feb 25 '11 at 16:31

score 6 · Answer 1 · answered Feb 25 '11 at 14:32

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.

score 5 · Accepted Answer · answered Feb 25 '11 at 14:20

5

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

answered Feb 25 '11 at 14:20

NebulousReveal

13,935

1

But why? ${ }{}{}{}$ – Pedro Jun 20 '12 at 01:54

score 1 · Answer 3 · answered Feb 25 '11 at 14:19

1

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

answered Feb 25 '11 at 14:19

lhf

216,483

score 1 · Answer 4 · answered Feb 25 '11 at 16:00

1

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

answered Feb 25 '11 at 16:00

yoyo

9,943

1

$\log_b(x)$ is not increasing for $0 \lt b \lt 1$ though it is still monotonic – Henry Mar 14 '12 at 00:18

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

4 Answers4

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