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

Aug 5, 2015

$x = \frac{1}{2} \text{ }$ or $\text{ } x = \ln \frac{3}{\ln} 4$

#### 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?"

${4}^{x} + 6 \left({4}^{-} x\right) = 5$

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:

${4}^{x} + \frac{6}{4} ^ x = 5$

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

${\left({4}^{x}\right)}^{2} + 6 = 5 \left({4}^{x}\right)$

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.

${\left({4}^{x}\right)}^{2} - 5 \left({4}^{x}\right) + 6 = 0 \text{ }$ Yes. We have:

${u}^{2} - 5 u + 6 = 0 \text{ }$ with ${4}^{x}$ rather than $u$.

$\left(u - 2\right) \left(u - 3\right) = 0$

$u = 2 \text{ }$ or $\text{ } u = 3$

${4}^{x} = 2 \text{ }$ or $\text{ } {4}^{x} = 3$

$x = \frac{1}{2} \text{ }$ or hmmmm. The best I can do is something like

$x = \frac{1}{2} \text{ }$ or $\text{ } x = {\log}_{4} 3$

If I'd rather have $\ln$ than ${\log}_{4}$, I'll write:

$x = \frac{1}{2} \text{ }$ or $\text{ } x = \ln \frac{3}{\ln} 4$