SOLUTION: How to solve these logarithmic equations: 1.) log3(x+6)-log3(x-2)=2 [with a base of 3] 2.) log8(x-6)+log8(x+6)=2 [with a base of 8] PLEASE HELP!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: How to solve these logarithmic equations: 1.) log3(x+6)-log3(x-2)=2 [with a base of 3] 2.) log8(x-6)+log8(x+6)=2 [with a base of 8] PLEASE HELP!       Log On


   

Question 769923: How to solve these logarithmic equations:
1.) log3(x+6)-log3(x-2)=2 [with a base of 3]
2.) log8(x-6)+log8(x+6)=2 [with a base of 8]
PLEASE HELP!
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
How to solve these logarithmic equations:
1.) log3(x+6)-log3(x-2)=2 [with a base of 3]
2.) log8(x-6)+log8(x+6)=2 [with a base of 8]
Ans:
Remember that Log a - log b = log (a/b)
log a + log b = log (a*b)
Also, if loga(x) = n i.e. log x to base a = n, then x = a^n

1)
logx(x+6) - log3(x-2) = log3((x+6)/(x-2)) = 2
So %28x+%2B+6%29%2F%28x+-+2%29+=+3%5E2+=+9
x+%2B+6+=+9%2A%28x+-+2%29+=+9%2Ax+-+18
Simplifying and bring like terms to one side,
8%2Ax+=+24 or
x+=+3

2)
log8(x-6)+log8(x+6)= log8((x - 6)*(x + 6)) = 2
%28x+-+6%29%2A%28x+%2B+6%29+=+8%5E2+=+64
x%5E2+-+36+=+64
x%5E2+=+100
x = 10 or x = -10

:)