SOLUTION: I will try to write this correctly - 5^x-1 = 125^2x+3 ( the x-1 and 2x+3 should all be in superscript and raised) HELP need to solve thank lucy

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I will try to write this correctly - 5^x-1 = 125^2x+3 ( the x-1 and 2x+3 should all be in superscript and raised) HELP need to solve thank lucy       Log On


   

Question 92951: I will try to write this correctly -
5^x-1 = 125^2x+3 ( the x-1 and 2x+3 should all be in superscript and raised)
HELP
need to solve
thank
lucy
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
5%5E%28x-1%29+=+125%5E%282x%2B3%29


5%5E%28x-1%29+=+%285%5E3%29%5E%282x%2B3%29 Rewrite 125 as 5%5E3

5%5E%28x-1%29+=+5%5E%283%282x%2B3%29%29 Multiply the exponents

5%5E%28x-1%29+=+5%5E%286x%2B9%29 Distribute


Now that we have the same base, we can set the exponents equal to each other


x-1=6x%2B9


x=6x%2B9%2B1 Add 1 to both sides


x-6x=9%2B1 Subtract 6x from both sides



-5+x=9%2B1 Combine like terms 1+x and -6+x on the left side to get -5+x




-5+x=10 Combine like terms 1 and 9 on the right side to get 10





x=%2810%29%2F%28-5%29 Now divide both sides by -5 to isolate and solve for x


x=+-2 Reduce



So our answer is
x=+-2


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Check:

5%5E%28-2-1%29+=+125%5E%282%28-2%29%2B3%29 Plug in x=+-2 into the original equation

5%5E%28-2-1%29+=+125%5E%28-4%2B3%29 Multiply

5%5E%28-3%29+=+125%5E%28-1%29 Combine like terms in the exponents

1%2F5%5E3+=+1%2F125 Rewrite 125%5E%28-1%29 as 1%2F125%5E1 which is just 1%2F125. Rewrite 5%5E%28-3%29 as 1%2F5%5E3

1%2F125+=+1%2F125 Evaluate 1%2F5%5E3 to get 1%2F125. Since the equations are equal, our answer is verified.