Solving Exponential Equations by Factoring

Q:  Solve for x:
2^2x + 2^x+2 – 12 = 0

A:  The first thing we need to notice (from practice and experience) is that we can re-write this like so using our knowledge / rules of exponents:

(2^x)² + 2²*2^x – 12 = 0
Now simplify a little:

(2^x)² + 4*2^x – 12 = 0

So, look at this in a new light.  What if we substitute each 2^x with y?

(2^x)² + 4*2^x – 12 = 0   turns into   y² + 4*y – 12 = 0

(this isn’t necessary, but is helpful for visualization)

We see this is in the form of a quadratic and can be factored:

(y + 6)(y – 2) = 0

So, y = -6 or y = 2

Remember, y was a substitution for 2^x. So, we really have:
2^x = -6   or   2^x = 2

Solve each equation separately.  Let’s start with:

2^x = -6

Solve for x by taking the log of both sides (you can use the log of any base: log base 10, log base 2, natural log):

log(2^x)= log(-6)
We can stop right here because you cannot take the log of a negative number. This equations yields no solutions.

So, the second equation:

2^x = 2
log(2^x) = log(2)

Logarithm rules say that the x exponent can come down as a multiplier like so:

x*log(2) = log(2)

Divide both sides by log(2) to get:

x = 1.