Question 35663
I am assuming that you mean {{{log(8,3x-1) = log(8,x+4) }}}


I have coined the phrase which I call the "This Equals That" Theorem:
"If {{{log(b,THIS)= log(b,THAT)}}}", 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(8,3x-1) = log(8,x+4) }}}, then {{{3x-1 = x+4}}}, provided none of the values of x causes a log of a negative!!


Solve for x:

{{{3x-1 = x+4}}}
{{{2x = 5}}}
{{{x= 5/2}}}


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