+33666 Take logs of both sides: ln(2x3x-1)=ln(5x-1) ln is the natural, or Napierian logarithm.
Using a property of logarithms (see the Formulary on this site) we can write ln(2x)+ln(3x-1) = ln(5x-1)
Using another property of logarithms (again, see the Formulary) this becomes x*ln(2) + (x-1)*ln(3) = (x-1)*ln(5)
Expand the bracketed terms and collect: x*(ln(5) - ln(2) - ln(3)) = ln(5) - ln(3) or x*ln(5/6) = ln(5/3) so x = ln(5/3)/ln(5/6)
$${\mathtt{x}} = {\frac{{ln}{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}}{{ln}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}}} = {\mathtt{x}} = -{\mathtt{2.801\: \!784\: \!016\: \!923\: \!930\: \!8}}$$
+33666 Take logs of both sides: ln(2x3x-1)=ln(5x-1) ln is the natural, or Napierian logarithm.
Using a property of logarithms (see the Formulary on this site) we can write ln(2x)+ln(3x-1) = ln(5x-1)
Using another property of logarithms (again, see the Formulary) this becomes x*ln(2) + (x-1)*ln(3) = (x-1)*ln(5)
Expand the bracketed terms and collect: x*(ln(5) - ln(2) - ln(3)) = ln(5) - ln(3) or x*ln(5/6) = ln(5/3) so x = ln(5/3)/ln(5/6)
$${\mathtt{x}} = {\frac{{ln}{\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right)}}{{ln}{\left({\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}}} = {\mathtt{x}} = -{\mathtt{2.801\: \!784\: \!016\: \!923\: \!930\: \!8}}$$
+118724 Thanks Alan
I just thought that i would point out that this does not have to be base e. It can be base 10 or base whatever you want. Calculators only deal with base e and base 10 so you had best use one of those. 
Sacha, if you have any problem understanding what Alan has done make sure your ask for clarification. We all love to get polite feedback. (positive, negative or queries)
