How to use log properly in this task?

thanks anon, that was a good start. :)

I am going to use     log base 10     because      $$log_{10}10=1$$

$$\\3^ {2n-1} < 10^ {n-1 }\\\\
Log_{10}3^ {2n-1} (2n-1)Log_{10}3<(n-1)Log_{10} 10\\\\
Log_{10}3(2n-1)<(n-1)\\\\
2nLog_{10}3-Log_{10}3 2nLog_{10}3-n n(2Log_{10}3-1)

$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}} = -{\mathtt{0.045\: \!757\: \!490\: \!560\: \!675\: \!1}}$$     

I am going to divide by this number and since it is negative I will have to turn the sign around.

$$\\n>\frac{log_{10}(3)-1}{2log_{10}(3)-1}$$

$${\frac{\left({log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{1}}\right)}} = {\mathtt{11.427\: \!172\: \!663\: \!391\: \!417\: \!6}}$$

$$n\ge 12 \qquad where \;\;n\in N$$  

here is my helpm you incorporate the log on the left and right and continue like so

I wanted to give you thumbs up anon but it won't let me. 

There seems to be some problems with the system tonight.