Solving Exponential Equations...

Oct 2012
11
3
London
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
May 2006
12,028
6,341
Lexington, MA (USA)
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?
 
Oct 2012
11
3
London
That was embarrasingly simple! I knew I was missing something elementary.

Many thanks Soroban!