SOLUTION: solve each equation. If the answer is not exact give the answer to 4 decimal places. 3^(x+3)+3^x=84 4^(x+2)-4^x=15

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: solve each equation. If the answer is not exact give the answer to 4 decimal places. 3^(x+3)+3^x=84 4^(x+2)-4^x=15      Log On


   

Question 384403: solve each equation. If the answer is not exact give the answer to 4 decimal places.
3^(x+3)+3^x=84
4^(x+2)-4^x=15
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
3%5E%28x%2B3%29%2B3%5Ex=84
One way to solve this involves factoring in a way you may not have seen/done before. The first term on the left is the product of x+3 3's. The second term is the product of x 3's. So both terms have x 3's in common. We can factor out those x 3's from each term. After we factor out the x 3's, the first term will still have 3 3's. The second term will just be a 1 since we factored out all of its 3's:
%283%5Ex%29%283%5E3%2B1%29=84
which simplifies to:
%283%5Ex%29%2828%29=84
Dividing both sides by 28 we get:
3%5Ex+=+3
So x = 1

4%5E%28x%2B2%29-4%5Ex=15
This problem can be solved in a similar way:
%284%5Ex%29%284%5E2-1%29=15
%284%5Ex%29%2815%29=15
4%5Ex=1
And so, if you remember how zero exponents work, x = 0.