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The Number e and the Natural Logarithm - Problem 2
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I want to talk about some properties of logs as they apply to the natural logarithm. If you remember, the definition of natural logarithm is; y equals lnx means x equals e to the y. And that means that y equals lnx and y equals e to the x of inverse functions. As a result of this inverse relationship, we have two identities.

First ln of e to the x equals x and second, e to the lnx equals x. Now the second identity is only true for x greater than 0 because that’s the domain of natural log of x. So this second one x greater than 0, but this one is true for all x.

Now let’s apply this on an example. The problem says to evaluate these four things lnv to the 4, that’s just a straight forward application of this identity here. lnv to the 4 would be 4. Ln of 1 over e³. We first have to write this as a single power of e. So that would be natural log of e to the -3. 1 over e³ is e to the -3 so lnv of e of the -3, is -3.

Same thing here I need to write this root e as a power of e, but that’s e to the 1/2. And so ln of e to the 1/2 is 1/2.

And finally and you may know the answer to this one but this is kind of e. If you express 1 as e to the 0, you see y the natural log of 1 is 0. Any log, log of any base of 1 is 0, but you can use this inverse property to show it in a kind of nice way.

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