The sum of $8^x$ and its reciprocal

Question

The sum of $2^x$ and its reciprocal is 3.

So this makes the value of $x$ equals to $log_2(3 ± √5) - 1$ by using quadratic and logarithm

The sum of $8^x$ and its reciprocal Is 18 by substituting the value of $x$. My question is, is there a way to solve this within 1 minute? Like a clever shortcut or you just need to brute force this to find the answer?

Answer 1

Yes. You are given that

$$a+\frac{1}{a}=3$$

with $a=2^x$ and you want to compute

$$a^3+\frac{1}{a^3}$$

Just notice that

$$a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)=3^3-3\cdot 3=18$$

Answer 2

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

Answer 3

Suppose that $8^x+8^{-x}=18$. Then, multiplying both sides by $8^x$, we get $8^{2x}+1=18 \cdot8^x$. Moving the product from the right hand side to the left hand side, we get $8^{2x}-18 \cdot 8^x+1=0$. Using the quadratic formula (after substituting $u=8^x$ to get $u^2-18u+1=0$) then gives $8^x=u=\frac{18 \pm \sqrt{324-4}}{2}=\frac{18 \pm \sqrt{320}}{2}=\frac{18 \pm \sqrt{64 \cdot 5}}{2}=\frac{18 \pm 8\sqrt{5}}{2}=9 \pm 4\sqrt{5}$. Finally, both of the solutions for $u$ are positive, so the two solutions for $x$ are $\mathrm{log}_8(9+4\sqrt{5})$ and $\mathrm{log}_8(9-4\sqrt{5})$.