Solving $8^{2x}-2\cdot8^x+1=0$

Question

$8^{2x}-2\cdot8^x+1=0$, I tried a lot of ways to solve this equation, like changing $8$ to $2^3$, or writing $2*8^x$ as $2*2^{3x}$ and then $2^{3x+1}$, but i'm not getting anywhere, i have the solution which is $x=0$, but i don't know how to get there. Thanks.

Answer 1

So you can let $8^x = a$

$$8^{2x} -2(8^x)+1 = 0 $$

$$ (8^x)^2 - 2(8^x) + 1 = 0$$ $$ a^2 -2a +1 = 0 $$ $$ (a-1)^2 = 0$$ $$a = 1 $$ $$ 8^x = 1 = 8^0$$

$$x = 0$$

Answer 2

let $8^x=y$. The equation becomes $y^2-2y+1=(y-1)^2=0$ and so $y-1=0$ and we get $8^x=1$ and hence $x=0$.

Answer 3

Hint:
Reduce your exponential equation to a quadratic equation via $$ y = 8^x$$ to get

$$ y^2 - 2y+1 = 0 .$$