How would I use change of base to solve (e^x + e^-x) = 3? I know normally you would just take the natural log of both sides of the equation, but I forget the natural log property when it comes to something like ln (e^x + e^-x) because I don't remember if you can separate that quantity into two natural logs or not?

Question

myininaya · Answer

Change of base? 
I would just multiply e^x on both sides and realize we have a equation in quadratic form

anonymous · Answer

The book is asking to solve for x, so even if I did do that, I'd still have the x in the exponent and would need to change the equation into logarithmic form, wouldn't I?

myininaya · Answer

you would solve for e^x first, then x.

myininaya · Answer

To solve 
$$au^2+bu+c=0 \text{ you may use  } u=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
u could be e^x or even a spaceship

myininaya · Answer

like this I will replace all the u's there with e^x

$$ae^{2x}+be^x+c=0$$
Then
$$e^x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Then you would solve for x by doing ln( ) of both sides
However you will need to also verify the inside of the ln( ) is positive

anonymous · Answer

Okay, I so if I multiply $$(e^x)(e^x + e^{-x})$$ It would end up just being $$e ^{2x}$$Right? And then just move the equation and take the natural log of both sides, blah, blah, blah. I think I got the rest! Thank you!

myininaya · Answer

you have two terms in the parenthesis
you only showed one term is the result of multiplying the above

myininaya · Answer

Distribute

myininaya · Answer

$$e^x(e^x+e^{-x})=e^x \cdot e^x +e^x \cdot e^{-x}=e^{x+x}+e^{x-x}$$

anonymous · Answer

That's what I mean, you're supposed to add exponents. Therefore adding x to -x would - OH, make it e^0, aka 1?

myininaya · Answer

yeah
so you have $$e^{2x}+1=3e^{x}$$
Which then you would subtract 3e^x on both sides 
then use quadratic formula as suggested above

anonymous · Answer

Okay, thanks!

myininaya · Answer

np