Solving Exponential Equations...

• Oct 30th 2012, 03:27 AM
cac2008
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 $3^2^x-9(3^x)=0$ that you substitute for the exponential e.g. $y=3^x$ and solve from there, but i've now been presented with:

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

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)
• Oct 30th 2012, 04:48 AM
Soroban
Re: Solving Exponential Equations...
Hello, cac2008!

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

Quote:

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

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

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

Got it?
• Oct 30th 2012, 05:14 AM
cac2008
Re: Solving Exponential Equations...
That was embarrasingly simple! I knew I was missing something elementary.

Many thanks Soroban!