You can use the laws of logarithms to simplify your equations into one containing a single log
If I take your first equation we can use the first law to say that .
log 2x = log 3+ log (x-1)
ln x + ln (x-2) - ln (x+4) = 0
I can not figure out how to separate the variable from the constants.
a\ log 2x = log 3+ log (x-1)
b\ log 3+ log (x-1) - log 2x
d\ log 3/2 (1-(1/x))
c\ log 3/2 + log (1-(1/x))
this approach doesn't seem to be going any place.
i think i'm trying to get a sort of zero (the input needs to be 1 not zero) of the rational function inside of the logarithm. clearly;
(3x-3)/2x=1 when x = 3.
but i can't come up with a generic way to evaluate for x with out just randomly inserting numbers or using a graphing program.
i'm sorry if that wasn't as obvious as i thought it was in the first place or if i am incorrect and am doing the whole thing wrong.