# Solving Exponential Equations...

#### cac2008

Guys and Gals,

I'm going over my old textbooks in preparation for going back to university, and one question has me totally stumped.

I understand when encountering problems like $$\displaystyle 3^2^x-9(3^x)=0$$ that you substitute for the exponential e.g. $$\displaystyle y=3^x$$ and solve from there, but i've now been presented with:

$$\displaystyle 2^2^x-2^x^+^1+1=0$$

Can anyone advise me how I go about solving this? I'll no doubt kick myself when I see it, but I just can't figure out what to do! (Headbang)

#### Soroban

MHF Hall of Honor
Hello, cac2008!

We have to "hammer" the equation into the proper form.

Solve for $$\displaystyle x\!:\;2^{2x}-2^{x+1}+1\:=\:0$$

We have: .$$\displaystyle 2^{2x} - 2\cdot2^x + 1 \:=\:0$$

Let $$\displaystyle y \,=\,2^x\!:\;\;y^2 - 2y + 1 \:=\:0$$

Got it?

#### cac2008

That was embarrasingly simple! I knew I was missing something elementary.

Many thanks Soroban!