Question
Let $$\left( {{x_0},{y_0}} \right)$$ be the solution of the following equations
$$\eqalign{
& {\left( {2x} \right)^{\ell n2}} = {\left( {3y} \right)^{\ell n3}} \cr
& \,\,\,\,{3^{\ell nx}} = {2^{\ell ny}} \cr
& {\text{Then }}\,{x_0}\,{\text{is}} \cr} $$
A.
$$\frac{1}{6}$$
B.
$$\frac{1}{3}$$
C.
$$\frac{1}{2}$$
D.
6
Answer :
$$\frac{1}{2}$$
Solution :
We have
$$\eqalign{
& {\left( {2x} \right)^{\ell n2}} = {\left( {3y} \right)^{\ell n3}} \cr
& \Rightarrow \,\,\ell n2.\,\,\ell n2x = \ell n3.\,\,\ell n3y \cr
& \Rightarrow \,\,\ell n2.\,\,\ell n2x = \ell n3.\left( {\ell n3 + \ell ny} \right)\,\,\,\,.....\left( 1 \right) \cr
& {\text{Also given }}{3^{\ell nx}} = {2^{\ell ny}} \cr
& \Rightarrow \,\,\ell nx.\,\,\ell n3 = \ell ny.\,\,\ell n2 \cr
& \Rightarrow \,\,\ell ny = \frac{{\ell nx.\,\,\ell n3}}{{\ell n2}} \cr} $$
Substituting this value of $${\ell ny}$$ in equation (1), we get
$$\eqalign{
& \ell n2.\,\,\ell n2x = \ell n3\left[ {\ell n3 + \frac{{\ell nx.\,\,\ell n3}}{{\ell n2}}} \right] \cr
& \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x = {\left( {\ell n3} \right)^2}\ell n2 + {\left( {\ell n3} \right)^2}\ell nx \cr
& \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x = {\left( {\ell n3} \right)^2}\left( {\ell n2 + \ell nx} \right) \cr
& \Rightarrow \,\,{\left( {\ell n2} \right)^2}\ell n2x - {\left( {\ell n3} \right)^2}\ell n2x = 0 \cr
& \Rightarrow \,\,\left[ {{{\left( {\ell n2} \right)}^2} - {{\left( {\ell n3} \right)}^2}} \right]\ell n2x = 0 \cr
& \Rightarrow \,\,\ell n2x = 0 \cr
& \Rightarrow \,\,2x = 1\,\,{\text{or }}x = \frac{1}{2} \cr} $$