Solving Exponential Equations...

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)

Re: Solving Exponential Equations...

Hello, cac2008!

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

Quote:

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?

Re: Solving Exponential Equations...

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

Many thanks Soroban!