SOLUTION: 2^2x+3 -9(2^x)+1=0

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Question 422305: 2^2x+3 -9(2^x)+1=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2%5E%282x%2B3%29+-9%282%5Ex%29%2B1=0
The easiest way I see to solve this problem is not especially easy. It requires a good understanding of exponents and quadratic equations. So we will start with a short review of a couple of important ideas:
  • The rule for exponents when multiplying is to add the exponents:
    a%5Ep%2Aa%5Eq+=+a%5E%28p%2Bq%29
    For example, 5%5E3%2A5%5Ex+=+5%5E%283%2Bx%29
    As with most things in math, this works in reverse, too. We can take an exponent that is a sum and rewrite it as a product. For example,
    7%5E%283x%2B10%29+=+7%5E%283x%29%2A7%5E10
  • The general form for quadratic equations is:
    ax%5E2%2Bbx%2Bc+=+0
    The exponent in the first term is a 2 and the exponent of the middle term is 1. In other words the exponent of the first term is twice the exponent of the middle term. These are simple ideas you probably already know. What is less well known is that any equation where the exponent of the first term is twice the exponent of the middle term can be solved as a quadratic equation. These are called equations of "quadratic form". For example:
    x%5E12+-+3x%5E6+%2B+2+=+0
    can be solved as a quadratic equation because 12 is twice 6. (You'll see how shortly as we solve your equation.)

So to solve your equation we will rewrite the first term, using a%5Ep%2Aa%5Eq+=+a%5E%28p%2Bq%29 in reverse, so that its exponent is twice the exponent of your middle term:
2%5E%282x%29%2A2%5E3+-9%282%5Ex%29%2B1=0
And since 2%5E3+=+8 this becomes:
2%5E%282x%29%2A8+-9%282%5Ex%29%2B1=0
or
8%2A2%5E%282x%29+-9%282%5Ex%29%2B1=0
We now have the equation in quadratic form. The first few times yousolve one of these, it can be helpful to use a temporary variable. Set it equal to the base and exponent of the middle term:
Let q+=+2%5Ex
Then q%5E2+=+%282%5Ex%29%5E2+=+2%5E%282x%29
Substituting these into our equation we get:
8q%5E2-9q%2B1+=+0
This is clearly a quadratic equation. To solve it we factor it of use the Quadratic Formula. This factors fairly easily:
(8q-1)(q-1) = 0
From the Zero Product Property we know that one of these factors must be zero. So:
8q-1 = 0 or q-1 = 0
Solving these we get:
q = 1/8 or q = 1
Of course we are not interested in solutions for q. We are interested in solutions for x. So now we substitute back in for q. (Remember, q was just a temporary variable).
2%5Ex+=+1%2F8 or 2%5Ex+=+1
Solving these for x we get:
x = -3 or x = 0
These are two solutions to your equation. Check them if you like.

Note: Once you have done a few of these quadratic form equations you will no longer need a temporary variable. You will see how to go directly from
8%2A2%5E%282x%29+-9%282%5Ex%29%2B1=0
to
%288%2A2%5Ex+-+1%29%282%5Ex+-+1%29+=+0
to
8%2A2%5Ex+-+1+=+0 or 2%5Ex+-+1+=+0
etc.