Exponential turning point q

**Detanon** · November 9th 2010, 09:58 AM

Find the co ordinates of the turning point of the curve
$y = e^2^x - 3e^x$
and determine whether this point is a local max or local min.

I've differentiated this and put it = to 0 and got
$2e^2^x - 3e^x$ is that right? In the answer it says that x is loge^3/2 how do I change the e to log?
Thanks

**Unknown008** · November 9th 2010, 10:05 AM

Yes, so far so good.

Equate to zero now.

$2e^{2x} - 3e^x=0$

If you want, you can make a substitution of the type $y = e^x$ to help you.

**Detanon** · November 9th 2010, 10:23 AM

$e^x (2e^x-3) = 0$
$e^x =0 (cant be zero?) 2e^x = 3$
$log2e^3 = x$
but in the answer it says $loge^3/2$

**Unknown008** · November 9th 2010, 10:28 AM

If you mean this:

$log_e \dfrac32$

Then yes, it's the right answer.

$e^x = \dfrac32$

$log\ e^x = log \dfrac32$

$xlog\ e = log \dfrac32$

$x = \dfrac{log\ \dfrac32}{log\ e}$

$x = log_e\ \dfrac32$

**Detanon** · November 9th 2010, 10:34 AM

How did you do the last 2 lines? dont get how e ends up on top.

**Unknown008** · November 9th 2010, 10:38 AM

I thought you were told this law...

You have:

$\dfrac{log_a \ b }{log_a\ c} = \log_c\ b$

If in the fraction, the two logs have the same base, then you can change the base like I just did. This also applies in the reverse way.

Here are some other ways to change the base:

$\dfrac{1}{log_a\ b} = log_b\ a$

If:

$a^b = c$

Then;

$b = log_a\ c$

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