Exponential and Logarithm Equations

**JohnJames** · March 30th 2011, 08:21 PM

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

**pickslides** · March 30th 2011, 08:26 PM

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

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

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

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

Can you solve this?

**JohnJames** · March 31st 2011, 02:09 AM

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

**JohnJames** · March 31st 2011, 02:21 AM

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

**HallsofIvy** · March 31st 2011, 04:54 AM

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!

**JohnJames** · March 31st 2011, 05:16 AM

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

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