Question

How do you solve `4^x + 6(4^-x) = 5`?

Answer 1

`x=1/2" "` or `" "x=ln3/ln4`

Explanation:

Don't be too concerned about what you should do to solve the problem.
Focus on what you could do and on the question, "Would that help?"

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

Well, perhaps I could write 6 as 4 to some power. But i can add 4 to a powe plus 4 to another power, so that doesn't seem helpful.

Think of something else we could do.
Negative exponents can be a nit trick, so let's rewrite:

`4^x + 6/4^x = 5`

I've had good luck in other problems with clearing fraction, so let's do that.

`(4^x)^2 + 6 = 5 (4^x)`

Does that help? It's not really clear yet, but I do see a square, and a constant and a constant times the thing that is squared. That sound like a quadratic to me.

`(4^x)^2 -5(4^x)+ 6 = 0" "` Yes. We have:

`u^2 - 5u+6=0" "` with `4^x` rather than `u`.

`(u-2)(u-3)= 0`

`u = 2" "` or `" "u=3`

`4^x = 2" "` or `" "4^x=3`

`x=1/2" "` or hmmmm. The best I can do is something like

`x=1/2" "` or `" "x=log_4 3`

If I'd rather have `ln` than `log_4`, I'll write:

`x=1/2" "` or `" "x=ln3/ln4`