Let ( x_0,y_0 ) be the solution of the following equations \beginmatrix ( 2x )^\ell n2\, = ( 3y )^\ell n3 3^\ell nx\, = 2^\ell ny \endmatrix Then x_0 is

Quadratic Equation and Inequalities PYQ — JEE Advanced 2011 · Shift 1 · Q28 with Solution

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Correct answer: C

We are given the system

(2x)^{\ln 2} = (3y)^{\ln 3}

3^{\ln x} = 2^{\ln y}

We need to find $x_0$ .

Solve the second equation first:

3^{\ln x} = 2^{\ln y}

Taking natural logarithm on both sides,

\ln x \cdot \ln 3 = \ln y \cdot \ln 2

So,

(\ln 3)(\ln x) = (\ln 2)(\ln y)

Now take natural logarithm of the first equation:

(2x)^{\ln 2} = (3y)^{\ln 3}

Taking $\ln$ on both sides,

(\ln 2)(\ln(2x)) = (\ln 3)(\ln(3y))

Using log properties,

(\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y)

That is,

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)(\ln y)

From step 2,

(\ln 3)(\ln x) = (\ln 2)(\ln y)

Multiply both sides by $\dfrac{\ln 2}{\ln 3}$ or rewrite as

(\ln 2)(\ln x) = \frac{(\ln 2)^2}{\ln 3} \ln y

But a simpler way is to express one variable in terms of the other.

From

(\ln 3)(\ln x) = (\ln 2)(\ln y)

we get

\ln y = \frac{\ln 3}{\ln 2} \ln x

Substitute into the first logged equation:

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)\left(\frac{\ln 3}{\ln 2}\ln x\right)

So,

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + \frac{(\ln 3)^2}{\ln 2}\ln x

Multiply throughout by $\ln 2$ :

(\ln 2)^3 + (\ln 2)^2 \ln x = (\ln 2)(\ln 3)^2 + (\ln 3)^2 \ln x

Rearrange:

\left((\ln 2)^2 - (\ln 3)^2\right)\ln x = (\ln 2)(\ln 3)^2 - (\ln 2)^3

Factor both sides:

\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)\ln x = \ln 2\left((\ln 3)^2 - (\ln 2)^2\right)

\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)\ln x = -\ln 2\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)

Since the common factor is nonzero, cancel it:

\ln x = -\ln 2

Hence,

x = e^{-\ln 2} = \frac{1}{2}

Therefore,

x_0 = \frac{1}{2}

So the correct option is:

\boxed{\text{C}}

More from Quadratic Equation and Inequalities

View written solutionFree

Correct answer: C

We are given the system

(2x)^{\ln 2} = (3y)^{\ln 3}

3^{\ln x} = 2^{\ln y}

We need to find $x_0$ .

Solve the second equation first:

3^{\ln x} = 2^{\ln y}

Taking natural logarithm on both sides,

\ln x \cdot \ln 3 = \ln y \cdot \ln 2

So,

(\ln 3)(\ln x) = (\ln 2)(\ln y)

Now take natural logarithm of the first equation:

(2x)^{\ln 2} = (3y)^{\ln 3}

Taking $\ln$ on both sides,

(\ln 2)(\ln(2x)) = (\ln 3)(\ln(3y))

Using log properties,

(\ln 2)(\ln 2 + \ln x) = (\ln 3)(\ln 3 + \ln y)

That is,

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)(\ln y)

From step 2,

(\ln 3)(\ln x) = (\ln 2)(\ln y)

Multiply both sides by $\dfrac{\ln 2}{\ln 3}$ or rewrite as

(\ln 2)(\ln x) = \frac{(\ln 2)^2}{\ln 3} \ln y

But a simpler way is to express one variable in terms of the other.

From

(\ln 3)(\ln x) = (\ln 2)(\ln y)

we get

\ln y = \frac{\ln 3}{\ln 2} \ln x

Substitute into the first logged equation:

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + (\ln 3)\left(\frac{\ln 3}{\ln 2}\ln x\right)

So,

(\ln 2)^2 + (\ln 2)(\ln x) = (\ln 3)^2 + \frac{(\ln 3)^2}{\ln 2}\ln x

Multiply throughout by $\ln 2$ :

(\ln 2)^3 + (\ln 2)^2 \ln x = (\ln 2)(\ln 3)^2 + (\ln 3)^2 \ln x

Rearrange:

\left((\ln 2)^2 - (\ln 3)^2\right)\ln x = (\ln 2)(\ln 3)^2 - (\ln 2)^3

Factor both sides:

\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)\ln x = \ln 2\left((\ln 3)^2 - (\ln 2)^2\right)

\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)\ln x = -\ln 2\left((\ln 2 - \ln 3)(\ln 2 + \ln 3)\right)

Since the common factor is nonzero, cancel it:

\ln x = -\ln 2

Hence,

x = e^{-\ln 2} = \frac{1}{2}

Therefore,

x_0 = \frac{1}{2}

So the correct option is:

\boxed{\text{C}}

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