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Question Number 163996 by mathlove last updated on 12/Jan/22

{ ((9^((a+1)/b) =125)),((5^(b/a) =3)) :} then faind the volve of ((25^b )/(2a^2 ))=?

$$\begin{cases}{\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\mathrm{125}}\\{\mathrm{5}^{\frac{{b}}{{a}}} =\mathrm{3}}\end{cases} \\ $$$${then}\:\:{faind}\:{the}\:{volve}\:{of}\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }=? \\ $$

Answered by nurtani last updated on 12/Jan/22

5^(b/a) = 3 ⇒ 3 = 5^(b/a) .....(1) 9^((a+1)/b) = 125 ⇔ (3^2 )^((((a+1)/b))) =5^3 ⇔ 3^((((2a+2)/b))) = 5^3 ......(2) subtitution (1) & (2) (5^(b/a) )^((((2a+2)/b))) = 5^3 ⇔ 5^((((2a+2)/a))) = 5^3 ⇔ ((2a+2)/a) = 3 ⇒ 2a+2 = 3a ⇒ a = 2 5^(b/a) = 3 ⇒ 5^(b/2) = 3 ⇒ (5^b )^(1/2) = 3 ⇒ 5^b = 9 ∴ ((25^b )/(2a^2 )) = (((5^2 )^b )/(2a^2 )) = (((5^b )^2 )/(2a^2 )) = (9^2 /(2(2)^2 )) = ((81)/8)

$$\mathrm{5}^{\frac{{b}}{{a}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{3}\:=\:\mathrm{5}^{\frac{{b}}{{a}}} .....\left(\mathrm{1}\right) \\ $$$$\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\:\mathrm{125}\:\Leftrightarrow\:\left(\mathrm{3}^{\mathrm{2}} \right)^{\left(\frac{{a}+\mathrm{1}}{{b}}\right)} =\mathrm{5}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{3}^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{{b}}\right)} =\:\mathrm{5}^{\mathrm{3}} ......\left(\mathrm{2}\right) \\ $$$${subtitution}\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{5}^{\frac{\cancel{{b}}}{{a}}} \right)^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{\cancel{{b}}}\right)} =\:\mathrm{5}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{5}^{\left(\frac{\mathrm{2}{a}+\mathrm{2}}{{a}}\right)} \:=\:\mathrm{5}^{\mathrm{3}} \\ $$$$\Leftrightarrow\:\frac{\mathrm{2}{a}+\mathrm{2}}{{a}}\:=\:\mathrm{3}\:\Rightarrow\:\mathrm{2}{a}+\mathrm{2}\:=\:\mathrm{3}{a}\:\Rightarrow\:{a}\:=\:\mathrm{2} \\ $$$$\mathrm{5}^{\frac{{b}}{{a}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{5}^{\frac{{b}}{\mathrm{2}}} =\:\mathrm{3}\:\Rightarrow\:\left(\mathrm{5}^{{b}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\mathrm{3}\:\Rightarrow\:\mathrm{5}^{{b}} \:=\:\mathrm{9} \\ $$$$ \\ $$$$\therefore\:\:\:\:\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{5}^{\mathrm{2}} \right)^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\left(\mathrm{5}^{{b}} \right)^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} }\:=\:\frac{\mathrm{9}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}\right)^{\mathrm{2}} }\:=\:\:\frac{\mathrm{81}}{\mathrm{8}} \\ $$

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