How Do You Solve the Equation 4^x + 6(4^-x) = 5?

[Acnhduy]

Homework Statement

4^x + 6(4^-x) = 5

Homework Equations

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

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 :)

 

[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?

 

[Acnhduy]

it is actually,

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

 

[Acnhduy]

Quadratic :) thankyou

 

[Simon Bridge]

Well done :)