JEE Main Logarithms Question | Change of Base Formula PYQ

❓ Question

If

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

find the value of

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

---

✍️ Solution

Using the change of base formula,

$$m=\frac{\log_5 175}{\log_5 245}.$$

Now,

$$175=5^2\cdot7,\qquad 245=5\cdot7^2.$$

Therefore,

$$m= \frac{2+\log_57}{1+2\log_57}.$$

Rearranging,

$$m(1+2\log_57)=2+\log_57.$$
$$(2m-1)\log_57=2-m.$$

Hence,

$$\boxed{\log_57=\frac{2-m}{2m-1}} \qquad (1)$$

---

Similarly,

$$n=\frac{\log_5875}{\log_51215}.$$

Now,

$$875=5^3\cdot7,\qquad 1215=5\cdot3^5.$$

Therefore,

$$n= \frac{3+\log_57}{1+3\log_57}.$$

Rearranging,

$$n(1+3\log_57)=3+\log_57.$$
$$(3n-1)\log_57=3-n.$$

Hence,

$$\boxed{\log_57=\frac{3-n}{3n-1}} \qquad (2)$$

---

Equating (1) and (2),

$$\frac{2-m}{2m-1} = \frac{3-n}{3n-1}.$$

Cross-multiplying,

$$(2-m)(3n-1) = (3-n)(2m-1).$$

Expanding,

$$6n-2-3mn+m = 6m-3-2mn+n.$$

Simplifying,

$$5(n-m)+1-mn=0.$$

Hence,

$$1-mn=5(m-n).$$

Therefore,

$$\boxed{\frac{m-n}{1-mn}=\frac15.}$$

---

✅ Final Answer

$$\boxed{\frac15}$$