Exponential equation problem with no solution?

Question

I have trouble solving this:

$$3^x+3^{2-x}=8$$

I have tried substituting $3^x=z$ but that doesn't seem to help much.

Answer 1

Hint:

Let me just rewrite the equation for you

$$3^x + 3^{2-x} = 8\\ 3^x + 3^2\cdot 3^{-x} = 8\\ 3^x + 9\cdot (3^x)^{-1} = 8.$$

Now, try the substitution again. What do you get?

Answer 2

You can also rewrite it as $$ 3^{2x} - 8\cdot3^x + 3^2 = 0 $$

given that $3^{-x} \ne 0$. Then, solving for $3^x$ gives you

$$ 3^x = \frac{4 \pm \sqrt{16 - 9}}{1}. $$

Answer 3

Multiply $3^x$ to both sides: $3^{2x} + 9 = 8\cdot 3^x \implies 3^{2x} - 8\cdot 3^x + 9 = 0\implies (3^x - 4)^2 = 7 \implies 3^x - 4 = \pm \sqrt{7}\implies 3^x = 4 \pm \sqrt{7} \implies x = \log_{3}(4 \pm \sqrt{7})$.