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Does ` log_e (x/y) = (log_e x)/(log_e y)` ?  Explain why or why not.

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lbawa | Student, Undergraduate | (Level 1) Honors

Posted January 28, 2012 at 10:56 AM via web

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Does ` log_e (x/y) = (log_e x)/(log_e y)` ?  Explain why or why not.

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted January 28, 2012 at 11:23 AM (Answer #1)

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That equation is generally not true. First off, the logarithm with a base of e has a special name. It is actually the "natural" logarithm:

`log_e x = lnx`

To see why these two aren't equal, let's take each side apart. Let's start with the left side and let it equal a:

`ln(x/y) = a`

We know that, based on the definition of logs, that this is equivalent to saying:

`e^a = x/y`

Let's now represent `x` by `e^b` and `y` by `e^c`. This representation would mean:

`lnx = b`

`ln y = c`

However, based on our equation above, this would also mean:

`e^a = x/y = e^b/e^c = e^(b-c)`

Just rewriting so we can simplify:

`e^a = e^(b-c)`

Now, let's take the natural log of both sides! (remember` ln(e^z) = z` based on the definition of logs!)

`lne^a = lne^(b-c)`

`a = b-c`

But now, remember what a, b, and c are! Let's substitute our values for these dummy variables!

`ln(x/y) = lnx - lny`

So, we have our answer for what `ln(x/y)` is! It is clearly not `lnx/lny`

So, this begs the question, what is `lnx/lny`?

Well, I could show you the derivation for this (see link for this), but just to speed things up suffice it to say that it is the "change of base" formula:

`log_y x = lnx/lny`

This is how you can enter any logarithm into a normal calculator. In fact, the change of base formula isn't limited to natural logs. It can look like:

`log_y x = (log_c x) / (log_c y)`

for any c!

So there you have it.

`ln(x/y) = lnx - lny`

and

`lnx/lny = log_y x`

Our answer officially, then is:

`ln(x/y) != lnx/lny`

Hope that helps!

Sources:

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