SOLUTION: 1.)Solve the equation 2^(2x – 1) = 8^(x + 7). 2.)Solve the equation 3^x = 3^(3x + 1). 3.)Solve the equation (1/3)^x = 3^(x – 6). I just can't figure these out. Thanks so much

Algebra ->  Exponents -> SOLUTION: 1.)Solve the equation 2^(2x – 1) = 8^(x + 7). 2.)Solve the equation 3^x = 3^(3x + 1). 3.)Solve the equation (1/3)^x = 3^(x – 6). I just can't figure these out. Thanks so much       Log On


   



Question 558011: 1.)Solve the equation 2^(2x – 1) = 8^(x + 7).
2.)Solve the equation 3^x = 3^(3x + 1).
3.)Solve the equation (1/3)^x = 3^(x – 6).
I just can't figure these out. Thanks so much for helping!!!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1.) 2%5E%282x-1%29+=+8%5E%28x%2B7%29
Since 8=2%5E3 , substituting in the equation given, we get
2%5E%282x-1%29+=+%282%5E3%29%5E%28x+%2B+7%29 and applying the properties of exponents, that simplifies to
2%5E%282x-1%29+=+2%5E%283%2A%28x+%2B+7%29%29 , which simplifies further to
2%5E%282x-1%29+=+2%5E%283x%2B21%29
Since both sides are powers of the same base (2), the exponents must be equal, so we solve
2x-1=3x%2B21 --> -1=x%2B21 --> -22=x
Checking: 2%5E%282x-1%29+=2%5E%282%28-22%29-1%29=2%5E%28-44-1%29=2%5E-45 and 8%5E%28x%2B7%29=8%5E%28-22%2B7%29=8%5E-15=%282%5E3%29%5E-15=2%5E%283%2A%28-15%29%29=2%5E-45
2.) 3%5Ex+=+3%5E%283x+%2B+1%29
Since both sides are powers of the same base (3), the exponents must be equal, so we solve
x=3x%2B1 --> 0=2x%2B1 --> 2x=-1 --> x=-1%2F2
3.) %281%2F3%29%5Ex+=+3%5E%28x-6%29
1%2F3=3%5E-1, so we can write the equation as
%283%5E-1%29%5Ex=+3%5E%28x-6%29 and applying the properties of exponents, that simplifies to
3%5E%28-x%29=+3%5E%28x-6%29
Since both sides are powers of the same base (3), the exponents must be equal, so we solve
-x=x-6 --> -x%2B6=x --> 6=x%2Bx --> 2x=6 --> x=6%2F2 --> x=3