Posted on

Solving Exponential Equations by Factoring

Q:  Solve for x:
22x + 2x+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:

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

(2x)2 + 4*2x – 12 = 0

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

(2x)2 + 4*2x – 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 2x. So, we really have:
2x = -6   or   2x = 2

Solve each equation separately.  Let’s start with:

Continue reading Solving Exponential Equations by Factoring

Posted on

Half-Life Word Problem

Q:  A scientist has 256 g of goo. After 195 minutes, her sample has decayed to 16 g. What’s the half-life of the goo in minutes? (Assume exponential decay)


First we need to set up the problem with our decay formula.  Depending on your school / teacher / book, the variables in the formulas may be different letters, but they all mean the same thing:

P = A*e^(rt)

P is the ending amount, A is the initial amount, r is the growth rate/decay, and t is time.

So, the set-up with our info:

16 = 256*e^(r*195)

Now we need to solve for r:

16 = 256*e^(r*195)

1/16 = e^(r*195)

Take the natural log of both sides: Continue reading Half-Life Word Problem