Solve $3^{2x} -2 \cdot 3^{x+5} + 3^{10} = 0$ for $x$

Question

Here's the question: Solve for $x$ in $$3^{2x} - 2 \cdot 3^{x+5} + 3^{10} = 0$$ I know that you have to factor something out, I'm just not sure what that something is. Thanks in advance

Answer 1

Hint: Substitute $y=3^x$ to get $$y^2-2\cdot 3^5\cdot y+3^{10}=0$$

Answer 2

HINT:

$$\left(\dfrac{3^x}{3^5}\right)^2-2\left(\dfrac{3^x}{3^5}\right)+1=0$$

$$\iff\left(\dfrac{3^x}{3^5}-1\right)^2=0$$

Now $3^x=3^5\iff3^{x-5}=1=3^0$

As $3\ne\pm1, x-5=0$

Answer 3

Use the substitution $u = 3^x$ to get $$u^2-2\cdot 3^5\cdot u+3^{10}= \left(u-3^5\right)^2 = 0$$ which is quadratic in $u$ and has a single real solution $u = 3^5$. So back-substituting yields $$3^x = 3^5 \iff x=5.$$