[tex]\bf \log_5(3)=a\qquad \qquad \log_5(4)=b\ \\\\\\ \log_5(4)=b\implies \log_5(2^2)=b\implies 2\log_5(2)=b\implies \log_5(2)=\cfrac{b}{2} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \log_{15}(2)\implies \stackrel{\textit{change of base rule}}{\cfrac{\log_5(2)}{\log_5(15)}}\implies \cfrac{~~\frac{b}{2}~~}{\log_5(3\cdot 5)}\implies \cfrac{~~\frac{b}{2}~~}{\log_5(3)+\log_5(5)} \\\\\\ \cfrac{~~\frac{b}{2}~~}{a+1}\implies \cfrac{b}{2}\cdot \cfrac{1}{a+1}\implies \cfrac{b}{2a+2}[/tex]
