# Thread: Exponential turning point q

1. ## Exponential turning point q

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

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

2. Yes, so far so good.

Equate to zero now.

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

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

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

4. If you mean this:

$\displaystyle log_e \dfrac32$

Then yes, it's the right answer.

$\displaystyle e^x = \dfrac32$

$\displaystyle log\ e^x = log \dfrac32$

$\displaystyle xlog\ e = log \dfrac32$

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

$\displaystyle x = log_e\ \dfrac32$

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

6. I thought you were told this law...

You have:

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

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:

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

If:

$\displaystyle a^b = c$

Then;

$\displaystyle b = log_a\ c$