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Question Number 145119 by imjagoll last updated on 02/Jul/21
(5^(log _(5/3) (5)) /3^(log _(5/3) (3)) ) =?
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} }\:=?\: \\ $$
Answered by Rasheed.Sindhi last updated on 03/Jul/21
(5^(log _(5/3) (5)) /3^(log _(5/3) (3)) ) =? log _(5/3) (5)=a ; log _(5/3) (3)=b ((5/3))^a =5 ; ((5/3))^b =3 ((((5/3))^a )/(((5/3))^b ))=(5/3) ((5/3))^(a−b) =((5/3))^1 a−b=1→b=a−1 (5^(log _(5/3) (5)) /3^(log _(5/3) (3)) )=(5^a /3^b )=(5^a /3^(a−1) ) =3((5/3))^a =3((5/3))^(log _(5/3) (5)) =3(5)=15 determinant (((b^(log_b (a)) =a)))
$$\:\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} }\:=?\: \\ $$$$\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)={a}\:\:\:;\:\:\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)={b} \\ $$$$\:\:\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{a}} =\mathrm{5}\:\:\:;\:\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{b}} =\mathrm{3} \\ $$$$\frac{\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{a}} }{\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{b}} }=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{a}−{b}} =\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{1}} \\ $$$${a}−{b}=\mathrm{1}\rightarrow{b}={a}−\mathrm{1} \\ $$$$\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} }=\frac{\mathrm{5}^{{a}} }{\mathrm{3}^{{b}} }=\frac{\mathrm{5}^{{a}} }{\mathrm{3}^{{a}−\mathrm{1}} } \\ $$$$=\mathrm{3}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{a}} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} =\mathrm{3}\left(\mathrm{5}\right)=\mathrm{15} \\ $$$$\:\: \\ $$$$\begin{array}{|c|}{\mathrm{b}^{\mathrm{log}_{\mathrm{b}} \left(\mathrm{a}\right)} =\mathrm{a}}\\\hline\end{array} \\ $$
Answered by liberty last updated on 03/Jul/21
let { ((log _(5/3) (5)=x ⇒5^(x−1) =3^x )),((log _(5/3) (3)=y⇒5^y =3^(y+1) )) :} { ((5 = 3^(x/(x−1)) )),(((3^(x/(x−1)) )^y = 3^(y+1) )) :}⇒((xy)/(x−1))=y+1 ⇒ xy=xy+x−y−1 ⇒x=y+1 . so the value of (5^(log _(5/3) (5)) /3^(log _(5/3) (3)) ) = (5^x /3^y ) = ((5.5^y )/3^y ) ⇒ 5×((5/3))^y = 5×3 = 15
$$\mathrm{let}\:\begin{cases}{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)=\mathrm{x}\:\Rightarrow\mathrm{5}^{{x}−\mathrm{1}} =\mathrm{3}^{{x}} }\\{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)={y}\Rightarrow\mathrm{5}^{{y}} =\mathrm{3}^{{y}+\mathrm{1}} }\end{cases} \\ $$$$\:\begin{cases}{\mathrm{5}\:=\:\mathrm{3}^{\frac{{x}}{{x}−\mathrm{1}}} }\\{\left(\mathrm{3}^{\frac{{x}}{{x}−\mathrm{1}}} \right)^{{y}} =\:\mathrm{3}^{{y}+\mathrm{1}} }\end{cases}\Rightarrow\frac{{xy}}{{x}−\mathrm{1}}={y}+\mathrm{1} \\ $$$$\Rightarrow\:{xy}={xy}+{x}−{y}−\mathrm{1} \\ $$$$\Rightarrow{x}={y}+\mathrm{1}\:.\:{so}\:{the}\:{value}\:{of} \\ $$$$\:\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} }\:=\:\frac{\mathrm{5}^{{x}} }{\mathrm{3}^{{y}} }\:=\:\frac{\mathrm{5}.\mathrm{5}^{{y}} }{\mathrm{3}^{{y}} } \\ $$$$\Rightarrow\:\mathrm{5}×\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{{y}} =\:\mathrm{5}×\mathrm{3}\:=\:\mathrm{15}\: \\ $$
Answered by Rasheed.Sindhi last updated on 03/Jul/21
(5^(log _(5/3) (5)) /3^(log _(5/3) (3)) ) =(( (5^(log _(5/3) (5)) /3^(log _(5/3) (5)) )×3^(log _(5/3) (5)) )/3^(log _(5/3) (3)) ) =((5/3))^(log _(5/3) (5)) ×3^(log _(5/3) (5)−log _(5/3) (3)) determinant (((b^(log_b a) =a))) =5×3^(log_(5/3) ((5/3))) =5×3^1 =15 determinant (((log_a a=1)))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} }\: \\ $$$$ \\ $$$$\:\:\:=\frac{\:\:\:\:\frac{\mathrm{5}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} }×\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} \:\:\:}{\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} } \\ $$$$ \\ $$$$=\left(\frac{\mathrm{5}}{\mathrm{3}}\right)^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)} ×\mathrm{3}^{\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{5}\right)−\mathrm{log}\:_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\mathrm{3}\right)} \\ $$$$\begin{array}{|c|}{\mathrm{b}^{\mathrm{log}_{\mathrm{b}} \mathrm{a}} =\mathrm{a}}\\\hline\end{array} \\ $$$$=\mathrm{5}×\mathrm{3}^{\mathrm{log}_{\frac{\mathrm{5}}{\mathrm{3}}} \left(\frac{\mathrm{5}}{\mathrm{3}}\right)} =\mathrm{5}×\mathrm{3}^{\mathrm{1}} =\mathrm{15} \\ $$$$\begin{array}{|c|}{\mathrm{log}_{\mathrm{a}} \mathrm{a}=\mathrm{1}}\\\hline\end{array} \\ $$

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