1. Oct 2, 2013

### Acnhduy

1. The problem statement, all variables and given/known data
4^x + 6(4^-x) = 5

2. Relevant equations
log??? since this is the unit we are doing, but i'm not sure if it applies.

3. The attempt at a solution

I was thinking of changing 4^x to a variable like 'a', but 4^-x is not the same so I can't replace that with 'a' as well.
Then I though 4^-x = 1/4 ^x
So
4^x + 6[(1/4)^x] = 5

and I'm lost.
thanks :)

2. Oct 2, 2013

### Simon Bridge

You have:

$$4^{x}+6(4^{-x})=5$$ ... is that right?

... and you want to put it into some form that you can integrate or differentiate or something?
You suspect it has something to do with taking a logarithm - so do I :), since $\log_4(4^x)=x$ hugely simplifies the problem.

Your intuition to put $a=4^x$ is a good one - with that substitution you get:
$$a+\frac{6}{a}=5$$ ... which is hard to think about, so put it in standard form.
Hint: multiply both sides by $a$.
What sort of equation is that?

Last edited: Oct 2, 2013
3. Oct 2, 2013

### Acnhduy

it is actually,

4^x + 6(1/4)^x = 5

4. Oct 2, 2013