SOLUTION: Find the exact solution, using common logarithms. 4^x+64(4^-x)=20

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Question 354654: Find the exact solution, using common logarithms.
4^x+64(4^-x)=20

Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!

Rewriting this with positive exponents we get:

Now we can eliminate the fraction by multiplying both sides by :

Using the Distributive Property on the left sides we get:

For the rule is to add the exponents. so . In the second term the 's cancel. So the equation is now:

Subtracting from each side we have:

Since the exponent on is twice the exponent of , then this equation is in quadratic form for . So we can solve it like a quadratic equation. We can factor the left side:

From the Zero Product Property we know that this (or any) product can be zero only if on of the factors is zero. So:
or
Adding 16 to both sides of the first equation and 4 to both sides of the second equation we get:
or
At this point we could mentally figure out the solutions. (What power of 4 is 16? and what power of 4 is 4? These are very simple questions to answer.) But since you were told to use common logarithms we will do so:
or
Now we can use the property of logarithms, to move the exponents in the arguments out front:
or
Then we can divide both sides of the both equations by log(4):
or
If you use your calculator on the logs in first equation and then divide them you should get 2 (or something extremely close to 2).

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