# Exponential turning point q

• Nov 9th 2010, 09:58 AM
Detanon
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
• Nov 9th 2010, 10:05 AM
Unknown008
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.
• Nov 9th 2010, 10:23 AM
Detanon
\$\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\$
• Nov 9th 2010, 10:28 AM
Unknown008
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\$
• Nov 9th 2010, 10:34 AM
Detanon
How did you do the last 2 lines? dont get how e ends up on top.
• Nov 9th 2010, 10:38 AM
Unknown008
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\$