**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} + 2^{2}*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 y^{2} + 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.

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