log question...easy one....

\hspace{-16}$Here $\bf{60^a=3}$ and $\bf{60^b=5\Leftrightarrow b=log_{60}(5)}$\\\\\\ and $\bf{60^{a+b}=15\Leftrightarrow (a+b)=\log_{60}(15)}$\\\\\\ So $\bf{1-(a+b)=\log_{60}(60)-\log_{60}(15)=\log_{60}\left(\frac{60}{15}\right)}$\\\\\\ So $\bf{1-(a+b)=\log_{60}(4)}$\\\\\\ and $\bf{(1-b)=1-log_{60}(5)=log_{60}(60)-log_{60}(5)=\log_{60}\left(\frac{60}{5}\right)}$\\\\\\ So $\bf{(1-b)=\log_{60}(12)}$\\\\\\ So $\bf{x=\frac{1-a-b}{2.(1-b)}=\frac{1}{2}.\frac{\log_{60}(4)}{\log_{60}(12)}=\frac{1}{2}.\log_{12}(4)=\log_{12}(2)}$\\\\\\ So $\bf{12^x=12^{\log_{12}(2)}=2}$