# Simple Calculus question

1. Apr 30, 2005

### erik05

Hello all. Really quick question here....What are the intervals of increase and decrease for y= e^x = e^-2x. I found the first derivative : y'= e^x-2e^-2x and set it equal to 0 but that's where I got stuck. How would you solve for x? I know that the answer is (ln2)/3 but how would you get there? Thanks.

2. Apr 30, 2005

### Pyrrhus

Working out the equation, and use some logarithm properties.

Apply natural logaritm

$$(e^{x})^{3} = 2$$

3. Apr 30, 2005

### erik05

Stupid question...but why to the exponent of 3?

4. Apr 30, 2005

### whozum

$$0= e^x-\frac{2}{e^{2x}}$$

Just swing that fraction to the other side and multiply out.

5. Apr 30, 2005

### erik05

I can't believe I didn't get that....thanks.