# Thread: Exponential and Logarithm Equations

1. ## Exponential and Logarithm Equations

Hello everybody,

I'm having trouble with this problem.

$3*2^x^-^2 = 6-2^x^-^2$

This homework question has me completely stumped, my teacher said that we should first solve for $2^x^-^2$ then take the log, but I have no clue what she means, and there is no similar problems in the book. I wish I would've asking her to clarify this. can anyone help me?

Thank You

2. $\displaystyle 3\times 2^{x-2} = 6-2^{x-2}$

$\displaystyle 3\times 2^{x-2} - 6+2^{x-2}= 0$

Make $u = 2^{x-2}$

$\displaystyle 3\times u - 6+u= 0$

Can you solve this?

3. Oh! it's the u substitution yes I can do that, sorry for the late response I went to sleep right after I posted. I will try to solve it now, Thank you

4. Originally Posted by pickslides
$\displaystyle 3\times 2^{x-2} = 6-2^{x-2}$

$\displaystyle 3\times 2^{x-2} - 6+2^{x-2}= 0$

Make $u = 2^{x-2}$

$\displaystyle 3\times u - 6+u= 0$

Can you solve this?
Actually now that I think of it, we only used u-substitution for integrals thats why I didn't think of that. Are you sure its not another way to do this?

Thank You

5. There are many different ways to solve equations but that is surely the simplest. I cannot imagine why you would not want to make an obvious substitution. Typically, one learns to solve equations with substitution, among other methods, several years before integration!

6. Originally Posted by HallsofIvy
There are many different ways to solve equations but that is surely the simplest. I cannot imagine why you would not want to make an obvious substitution. Typically, one learns to solve equations with substitution, among other methods, several years before integration!
Yes, I figured out how to do it shortly after posting that message.

$\displaystyle 3u +u= 6$

$\displaystyle u(3 +1)= 6$

$\displaystyle u(4)= 6$

$\displaystyle u= 6/4$

Then I plug the $2^x^-^2$ back in and use log.

Quite easy, I just didn't see it until pickslides pointed it out. Sometimes thats how math works LOL

Thank You