Is the logarithm function injective (one-to-one)?

Question

Is the logarithm function injective (or, one-to-one)?

In other words, does $\log_2(x) = \log_2(y) \implies x = y$?

I.e., as $x$ and $y$ are in the same log base, can I just drop the logs?

Thanks!

possible duplicate of Does $\log _b \left( x \right) = \log _b \left( y \right) \rightarrow x = y$ ? — Jonas Meyer, May 22 '11 at 18:54
You asked a question of which this is clearly a special case in February. Is there something that wasn't cleared up by the answers there? — Jonas Meyer, May 22 '11 at 19:05

score 2 · Accepted Answer · answered May 22 '11 at 14:21

2

What does it mean for a number $a$ to be equal to $\log_2(x)$? It means that $2^a=x$.

Can you use this to answer the question?

answered May 22 '11 at 14:21

Ah yes thank you, so $2^a = x = y$ hence $x = y$ correct? – Danny King May 22 '11 at 14:31
1

More like $2^a = 2^b$ and thus $2^{a-b} = 1$. – M.B. May 22 '11 at 14:33
2

More like $a=b$ and thus $2^a=2^b$. – GEdgar May 22 '11 at 19:04

1 Answers1