SOLUTION: Solve the equation log^8(3x-1)=log^8(x+4). Thank you!

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Question 35663: Solve the equation log^8(3x-1)=log^8(x+4).
Thank you!
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
I am assuming that you mean log%288%2C3x-1%29+=+log%288%2Cx%2B4%29+

I have coined the phrase which I call the "This Equals That" Theorem:
"If log%28b%2CTHIS%29=+log%28b%2CTHAT%29", then "THIS+=THAT"!!

(Someday, maybe I'll become famous for this theorem. Just remember, you heard it first HERE on algebra.com!!)

So, if log%288%2C3x-1%29+=+log%288%2Cx%2B4%29+, then 3x-1+=+x%2B4, provided none of the values of x causes a log of a negative!!

Solve for x:
3x-1+=+x%2B4
2x+=+5
x=+5%2F2

Check answers to make sure that you didn't accidentally have a log of a negative. Both logarithms are acceptable, so this is the final answer.

R^2 at SCC