Question Number 113426 by bemath last updated on 13/Sep/20
Answered by bemath last updated on 13/Sep/20
$$\left(\mathrm{1}\right)\:\mathrm{3}^{\mathrm{3}{b}} \:=\:\mathrm{5}^{\mathrm{4}{a}} \:\rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}^{\mathrm{3}{b}} \right)=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{4}{a}} \right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{3}^{\mathrm{2}{b}+\mathrm{2}} \:=\:\mathrm{5}^{\mathrm{3}{a}} \rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}^{\mathrm{2}{b}+\mathrm{2}} \right)=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{3}{a}} \right) \\ $$$$\begin{cases}{\mathrm{3}{b}=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{4}{a}} \right)}\\{\mathrm{2}{b}+\mathrm{2}=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{3}{a}} \right)}\end{cases} \\ $$$$\rightarrow\mathrm{2}\left(\frac{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{4}{a}} \right)}{\mathrm{3}}\right)+\mathrm{2}=\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{3}{a}} \right) \\ $$$$\rightarrow\mathrm{2}\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{4}{a}} \right)\right)=\mathrm{3log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{3}{a}} \right)−\mathrm{6} \\ $$$$\rightarrow\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{5}^{\mathrm{8}{a}} \right)=\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{5}^{\mathrm{9}{a}} }{\mathrm{3}^{\mathrm{6}} }\right) \\ $$$$\rightarrow\mathrm{3}^{\mathrm{6}} \:=\:\mathrm{5}^{{a}} \:.\:{Now}\: \\ $$$$\rightarrow\mathrm{9}^{{b}+\mathrm{1}} \:=\:\mathrm{125}^{{a}} \:=\:\left(\mathrm{5}^{{a}} \right)^{\mathrm{3}} \\ $$$$\rightarrow\:\mathrm{3}^{\mathrm{2}{b}+\mathrm{2}} \:=\:\left(\mathrm{3}^{\mathrm{6}} \right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{18}} \\ $$$$\rightarrow\:\mathrm{2}{b}+\mathrm{2}\:=\:\mathrm{18}\:;\:{b}\:=\:\mathrm{8}\: \\ $$$${and}\:\mathrm{9}^{\mathrm{9}} \:=\:\left(\mathrm{5}^{\mathrm{3}{a}} \right)\rightarrow\mathrm{3}{a}\mathrm{ln}\:\left(\mathrm{5}\right)=\mathrm{18ln}\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow{a}\:=\:\frac{\mathrm{6}\:\mathrm{ln}\:\left(\mathrm{3}\right)}{\mathrm{ln}\:\left(\mathrm{5}\right)}\:=\:\mathrm{6}.\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{3}\right) \\ $$