How do you solve this problem leaving logs in there?
$\displaystyle (7x+1)\ln(6)=(6x-1)\ln(7)$
Put all logs on a side, and the $\displaystyle x$ terms on the other side (distribute the ln's, that is) :
$\displaystyle \frac{7x+1}{6x - 1} = \frac{\ln(7)}{\ln(6)}$
Say $\displaystyle a = \frac{\ln(7)}{\ln(6)}$ (for ease of manipulation and handling) :
$\displaystyle \frac{7x + 1}{6x - 1} = a$
Get rid of the fraction by multiplying each side by $\displaystyle (6x - 1)$.
$\displaystyle 7x + 1 = a(6x - 1)$
Then expand and solve for $\displaystyle x$ (then substitute for $\displaystyle a$)
Does it help ?