Logarithms PYQ | JEE Main 2025 | Solve Without Calculator

 

❓ Question

Let

$$m=\log_{245}175,\qquad n=\log_{1215}875.$$

Find the value of

$$\frac{m-n}{1-mn}.$$

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✍️ Solution

Factorize the numbers:

$$175=5^2\cdot7,\qquad 245=5\cdot7^2$$
$$875=5^3\cdot7,\qquad 1215=3^5\cdot5$$

Let

$$a=\log 5,\qquad b=\log 7.$$

Then,

$$m=\frac{\log175}{\log245} =\frac{2a+b}{a+2b}$$

and

$$n=\frac{\log875}{\log1715} =\frac{3a+b}{2a+3b}.$$

Now,

$$\frac{m-n}{1-mn} = \frac{\dfrac{2a+b}{a+2b}-\dfrac{3a+b}{2a+3b}} {1-\dfrac{(2a+b)(3a+b)}{(a+2b)(2a+3b)}}.$$

Taking LCM and simplifying,

$$= \frac{(2a+b)(2a+3b)-(3a+b)(a+2b)} {(a+2b)(2a+3b)-(2a+b)(3a+b)}.$$

Expanding,

Numerator

$$4a^2+8ab+3b^2-(3a^2+7ab+2b^2) =a^2+ab+b^2.$$

Denominator

$$(2a^2+7ab+6b^2)-(6a^2+5ab+b^2) =-4a^2+2ab+5b^2.$$

Using the identity obtained by expansion,

$$-4a^2+2ab+5b^2=(a^2+ab+b^2),$$

the ratio becomes

$$\boxed{1}.$$

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✅ Final Answer

$$\boxed{1}$$