Question

If ${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$, ${3^{\ln x}} = {2^{\ln y}}$ and $({x_0},{y_0})$ is the solution of these equations, then ${x_0}$ is:
A) $\dfrac{1}{6}$
B) $\dfrac{1}{3}$
C) $\dfrac{1}{2}$
D) $6$

Answer 1

Hint: According to given in the question we have to determine the value of ${x_0}$ when ${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$and $({x_0},{y_0})$is the solution of these equations. So, first of all we have to take the ln in the both sides of the given expression ${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$
Now, to solve the expression obtained after taking ln in the both sides of the expression we have to use the formula for log as mentioned below:

Formula used: $ \ln (ab) = \ln a + \ln b................(A)$
Now, to solve ${3^{\ln x}} = {2^{\ln y}}$we have to take ln in the both sides of the expression and to simplify the expression obtained we have to apply cross-multiplication.
Now, we have to solve both of the obtained expressions to obtain the value ${x_0}$.

Complete step-by-step answer:
Step 1: First of all to solve the expression ${(2x)^{\ln 2}} = {(3y)^{\ln 3}}$we have to take log in the both sides of the expression as mentioned in the solution hint. Hence,
$ \Rightarrow (\ln 2)\ln (2x) = (\ln 3)(\ln 3y)$………………..(1)
Step 2: Now, to solve the expression (1) as obtained in the solution step 1, we have to use the formula (A) as mentioned in the solution hint. Hence,
$ \Rightarrow (\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y)$
On taking lny as a common term from the both sides of the expression obtained just above,
$
\Rightarrow \dfrac{{(\ln 2)}}{{(\ln y)}}(\ln 2 + \ln x) = (\ln 3)\left( {1 + \dfrac{{\ln 3}}{{\ln y}}} \right) \\
\Rightarrow (\ln 2)\left( {\dfrac{{\ln 2}}{{\ln y}} + \dfrac{{\ln x}}{{\ln y}}} \right) = \ln 3\left( {1 + \dfrac{{\ln 3}}{{\ln y}}} \right).........(2)
 $
Step 3: Now, we have to take ln on both sides of the expression ${3^{\ln x}} = {2^{\ln y}}$ as mentioned in the solution hint. Hence,
$ \Rightarrow (\ln x)(\ln 3) = (\ln y)(\ln 2)$
Now, to solve the expression obtained just above we have to apply cross-multiplication,
$ \Rightarrow \dfrac{{\ln x}}{{\ln y}} = \dfrac{{\ln 2}}{{\ln 3}}..................(2)$
Step 4: Now, to obtain the values ${x_0}$ and ${y_0}$ we have to solve the equations (1), and (2) as obtained in the solution steps 2, and 3.
$ \Rightarrow (\ln 2)\left( {\dfrac{{\ln 2}}{{\ln y}} + \dfrac{{\ln x}}{{\ln y}}} \right) = \ln 3\left( {1 + \dfrac{{\ln 3}}{{\ln y}}} \right)$
Which is the expression (1) just above, hence, on solving the expression,
$
\Rightarrow \dfrac{{{{(\ln 2)}^2} - {{(\ln 3)}^2}}}{{\ln y}} = \dfrac{{{{(\ln 3)}^2}{{(\ln 2)}^2}}}{{\ln 3}} \\
\Rightarrow \ln y = - \ln 3 \\
\Rightarrow \ln y = \dfrac{1}{{\ln 3}}
 $
On eliminating ln from the both sides of the expression,
$ \Rightarrow y = \dfrac{1}{3}$
Step 5: Now, with the help of expression (1) we can obtain the value of ${x_0}$same as the solution step 4.
$
   \Rightarrow \ln x = - \ln 2 \\
   \Rightarrow \ln x = \dfrac{1}{{\ln 2}}
 $
On eliminating ln from the both sides of the expression,
$ \Rightarrow x = \dfrac{1}{2}$
Final solution: Hence, with the help of the formula (A) as mentioned in the solution hint we have obtained the value of ${x_0} = \dfrac{1}{2}$.

Therefore option (C) is correct.

Note: If two terms of log are written in such a way that $\ln a + \ln b$ which is the addition of two log terms then with the help of properties of log we can represents the both terms as $\ln a \times \ln b$ which is the multiplication of two given terms of ln.
If two terms of log are written in such a way that $\ln a - \ln b$ which is the subtraction of two log terms then with the help of properties of log we can represents the both terms as $\dfrac{{\ln a}}{{\ln b}}$ which is the division of two given terms of ln.