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![\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}](/tpl/images/0007/1246/c23af.png)
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If [tex] log_{5}(3) = a[/tex]and [tex] log_{5}(4)= b[/tex]then [tex] log_{15}(2) [/tex]is? a)2b/a+1b...
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