SOLUTION: I have an exponential equation - not sure if I went about it in the right way or not... (125/8)^x+1 = (2/5)^x-1 for want of a better way of doing things I decided to ma

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Question 160268: I have an exponential equation - not sure if I went about it in the right way or not...
(125/8)^x+1 = (2/5)^x-1

for want of a better way of doing things I decided to make both values inside the parentheses to be divided by the same thing thus
(625/40)^x+1 = (16/40)^x-1
then decided to combine to get the exponentials from x+1 - (x-1) to = 2
and then combining the other to give me a final answer of (609/40)^2
Is this correct or how else should I do this please?
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

%28125%2F8%29%5E%28x%2B1%29+=+%282%2F5%29%5E%28x-1%29

Notice that 125=5%5E3 and that 8=2%5E3 so 125%2F8=%285%2F2%29%5E3
so replace 125%2F8 by %285%2F2%29%5E3:

%28%285%2F2%29%5E3%29%5E%28x%2B1%29+=+%282%2F5%29%5E%28x-1%29

Multiply exponents on the left:

%285%2F2%29%5E%283%28x%2B1%29%29+=+%282%2F5%29%5E%28x-1%29

Distribute: 

%285%2F2%29%5E%283x%2B3%29+=+%282%2F5%29%5E%28x-1%29

Now observe that on the right the base 2%2F5 is the 
reciprocal of the base on the left, and the reciprocal
is the -1 power, so write 2%2F5 on the right as %285%2F2%29%5E%28-1%29:

%285%2F2%29%5E%283x%2B3%29+=+%28%285%2F2%29%5E%28-1%29%29%5E%28x-1%29

Multiply the exponents on the right:

%285%2F2%29%5E%283x%2B3%29+=+%285%2F2%29%5E%28-1%28x-1%29%29

Distribute:

%285%2F2%29%5E%283x%2B3%29+=+%285%2F2%29%5E%28-x%2B1%29

Since both sides are powers of the same base, and
since that base is neither 0 nor 1, we may equate
the exponents and drop the bases:

3x%2B3=-x%2B1

4x=-2

x=-2%2F4

x=-1%2F2

Edwin