Expert: Paul Klarreich - 2/16/2008

Question
find the possible values of x for which
2^2x+1 = 3(2^x)-1

i got up to:
(2x+1)log2 = 3(2^x)-1
2xlog2 +log2 = 3(2^x)-1

then i got stuck...
i think i've started it wrong and i'm not really sure what to do with the 3(2^x)-1
how do i make it into the form log something?
any help would be greatly appreciated.
thanks

Answer
Questioner:   charlotte
Category:  Calculus
Private:  No
 
Subject:  logarithms
Question:  find the possible values of x for which
2^2x+1 = 3(2^x)-1

i got up to:
(2x+1)log2 = 3(2^x)-1
2xlog2 +log2 = 3(2^x)-1

then i got stuck...
i think i've started it wrong and i'm not really sure what to do with the 3(2^x)-1
how do i make it into the form log something?
any help would be greatly appreciated.
thanks
.............................
Hi, Charlotte,

You can't really apply logs when you have a sum -- it just goes nowhere.

I am going to assume that when you wrote  2^2x + 1, you meant  2^(2x) + 1, NOT 2^(2x + 1),
which seems likely.  So your equation is:

2^(2x) + 1 = 3(2^x) - 1

which is really nothing but a quadratic equation.

Remember that   2^(2x) =  (2^x)^2.  [Rule for product of exponents.]

(2^x)^2 + 1 = 3(2^x) - 1

Let y = 2^x, and your equation is:

y^2 + 1 = 3y - 1

y^2 - 3y + 2 = 0

(y - 2)(y - 1) = 0

y = 2  and y = 1, or
2^x = 2 and 2^x = 1

2^x = 2 = 2^1,  so  x = 1

2^x = 1 = 2^0,  so  x = 0

Them's your two solutions.