Logarithms & Exponents PYQ | JEE Main 2025 | Smart Shortcut

 

❓ Question

Given

$$2^{x+y}=6^y,\qquad 3^{x-1}=2^{y+1},$$

where

$$x=x_1,\qquad y=x_2.$$

Find

$$\frac{\log3-\log2}{x-y}.$$

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✍️ Solution

From

$$2^{x+y}=6^y=2^y\cdot3^y,$$

taking logarithm on both sides,

$$(x+y)\log2=y(\log2+\log3).$$

Expanding,

$$x\log2+y\log2=y\log2+y\log3.$$

Hence,

$$x\log2=y\log3,$$

or

$$\boxed{x=\frac{y\log3}{\log2}}.$$

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Now use the second equation,

$$3^{x-1}=2^{y+1}.$$

Taking logarithm,

$$(x-1)\log3=(y+1)\log2.$$

Substitute

$$x=\frac{y\log3}{\log2},$$

to get

$$\left(\frac{y\log3}{\log2}-1\right)\log3=(y+1)\log2.$$

Expanding,

$$y\frac{(\log3)^2}{\log2}-\log3 =y\log2+\log2.$$

Rearranging,

$$y\left(\frac{(\log3)^2-(\log2)^2}{\log2}\right) =\log2+\log3.$$

Using

$$a^2-b^2=(a-b)(a+b),$$
$$y\cdot \frac{(\log3-\log2)(\log3+\log2)}{\log2} =\log3+\log2.$$

Cancelling $(\log3+\log2)$,

$$\boxed{ y=\frac{\log2}{\log3-\log2} }.$$

Therefore,

$$x=\frac{y\log3}{\log2} =\frac{\log3}{\log3-\log2}.$$

Hence,

$$x-y = \frac{\log3-\log2}{\log3-\log2} =1.$$

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✅ Final Answer

$$\boxed{ \frac{\log3-\log2}{x-y} =\log3-\log2 =\log\!\left(\frac32\right) }$$