Logarithms PYQ | JEE Main 2025 | Smart Algebraic Trick

 

❓ Question

If

$$\frac{u+\log_{4}3}{u+\log_{2}3} = \frac{u+\log_{8}3}{u+\log_{4}3} =V,$$

then find the value of $V$.

---

✍️ Solution

Using the logarithm property,

$$\log_{a^n}b=\frac{1}{n}\log_ab,$$

let

$$t=\log_23.$$

Then,

$$\log_43=\frac{t}{2}, \qquad \log_83=\frac{t}{3}.$$

Hence,

$$\frac{u+\frac t2}{u+t} = \frac{u+\frac t3}{u+\frac t2}.$$

Cross-multiplying,

$$\left(u+\frac t2\right)^2 = \left(u+t\right)\left(u+\frac t3\right).$$

Expanding,

$$u^2+ut+\frac{t^2}{4} = u^2+\frac{4ut}{3}+\frac{t^2}{3}.$$

Simplifying,

$$ut-\frac{4ut}{3} = \frac{t^2}{3}-\frac{t^2}{4}.$$
$$-\frac{ut}{3} = \frac{t^2}{12}.$$

Therefore,

$$ut=-\frac{t^2}{4}$$

and since $t\neq0$,

$$\boxed{u=-\frac t4.}$$

Now,

$$V= \frac{u+\frac t2}{u+t} = \frac{-\frac t4+\frac t2}{-\frac t4+t}.$$
$$V = \frac{\frac t4}{\frac{3t}{4}} = \frac13.$$

---

✅ Final Answer

$$\boxed{\frac13}$$