SOLUTION: Find the unknown in the equation log5(-2500) = log5(5^n+2 minus 5^n+3) (Note: In one of the steps, you will need to get rid of the minus sign in -2500) Thanks for answering the

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find the unknown in the equation log5(-2500) = log5(5^n+2 minus 5^n+3) (Note: In one of the steps, you will need to get rid of the minus sign in -2500) Thanks for answering the       Log On


   

Question 133438: Find the unknown in the equation log5(-2500) = log5(5^n+2 minus 5^n+3) (Note: In one of the steps, you will need to get rid of the minus sign in -2500)
Thanks for answering the quesiton:) but how can I answer this question with logs appearing in at lest one of the steps(this is my teachers rule)
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the logs are equal so the quantities are equal

-2500=5^(n+2)-5^(n+3)

5^(n+3)=5^(n+2)(5^1)=5(5^(n+2))

substituting __ -2500=5^(n+2)-5(5^(n+2)) __ -2500=-4(5^(n+2))

dividing by -4 __ 625=5^(n+2)

taking log __ log(625)=(n+2)(log(5)) __ dividing by log(5) __ (log(625))/(log(5))=n+2

4=n+2 __ subtracting 2 __