Consider the function $\phi(x)=\ln x$ from $\mathbb R^+$ (under mult) to the group $\mathbb R$ (under add) . Prove that $\phi$ is an isomorphism.

Asked Apr 15 '24 at 17:20

Active Apr 15 '24 at 17:21

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Okay so I've proved all of the parts needed for an isomorphism except to show that $\phi(x)$ is onto. How do I go about doing that? I know that you start by assuming that $y \in \mathbb{R^+}$ and that we want to end up proving that for each of those ys there exists and x in \mathbb{R}. But how do I get there?

edited Apr 15 '24 at 17:21

asked Apr 15 '24 at 17:20

Whole Milk Slammer

5

Hint. Given $y=\ln x$, can you solve for $x$? – GEdgar Apr 15 '24 at 17:24
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Don't abbreviate words. Under mult? Under add? The internet is big enough for full words, and your goal should be not to confuse the volunteers you are asking help from. – Thomas Andrews Apr 15 '24 at 17:24
Many facts about the natural logarithm are easily deduced from facts about its inverse function $e^x$. – hardmath Apr 15 '24 at 17:32
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This has been asked before, I bet. – Shaun Apr 15 '24 at 18:04
IIRC this is a exercise in Rotman's group theory book. – person Apr 15 '24 at 18:13

Consider the function $\phi(x)=\ln x$ from $\mathbb R^+$ (under mult) to the group $\mathbb R$ (under add) . Prove that $\phi$ is an isomorphism.

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