Question Number 163996 by mathlove last updated on 12/Jan/22
$$\begin{cases}{\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\mathrm{125}}\\{\mathrm{5}^{\frac{{b}}{{a}}} =\mathrm{3}}\end{cases} \\ $$$${then}\:\:{faind}\:{the}\:{volve}\:{of}\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }=? \\ $$
Answered by nurtani last updated on 12/Jan/22
$$\mathrm{5}^{\frac{{b}}{{a}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{3}\:=\:\mathrm{5}^{\frac{{b}}{{a}}} …..\left(\mathrm{1}\right) \\ $$$$\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\:\mathrm{125}\:\Leftrightarrow\:\left(\mathrm{3}^{\mathrm{2}} \right)^{\left(\frac{{a}+\mathrm{1}}{{b}}\right)} =\mathrm{5}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{3}^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{{b}}\right)} =\:\mathrm{5}^{\mathrm{3}} ……\left(\mathrm{2}\right) \\ $$$${subtitution}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{5}^{\frac{\cancel{{b}}}{{a}}} \right)^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{\cancel{{b}}}\right)} =\:\mathrm{5}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{5}^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{{a}}\right)} \:=\:\mathrm{5}^{\mathrm{3}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{2}{a}+\mathrm{2}}{{a}}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{2}{a}+\mathrm{2}\:=\:\mathrm{3}{a}\:\Rightarrow\:{a}\:=\:\mathrm{2} \\ $$$$\mathrm{5}^{\frac{{b}}{{a}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{5}^{\frac{{b}}{\mathrm{2}}} =\:\mathrm{3}\:\Rightarrow\:\left(\mathrm{5}^{{b}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{5}^{{b}} \:=\:\mathrm{9} \\ $$$$ \\ $$$$\therefore\:\:\:\:\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{5}^{\mathrm{2}} \right)^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{5}^{{b}} \right)^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} }\:=\:\:\frac{\mathrm{81}}{\mathrm{8}} \\ $$