Logarithms PYQ | JEE Main 2025 | Telescoping Trick in 2 Minutes

 

❓ Question

Given that

$$x>13,$$

solve

$$\log_{x-1}1\cdot \log_{x-2}(x-1)\cdot \log_{x-3}(x-2)\cdots \log_{x-12}(x-11)=2.$$

Find the value of $x$.

---

✍️ Solution

Using the change of base formula,

$$\log_b a=\frac{\log a}{\log b},$$

the given expression becomes

$$\frac{\log1}{\log(x-1)} \cdot \frac{\log(x-1)}{\log(x-2)} \cdot \frac{\log(x-2)}{\log(x-3)} \cdots \frac{\log(x-11)}{\log(x-12)} =2.$$

The terms cancel telescopically:

$$\frac{\log x}{\log(x-12)}=2.$$

Using natural logarithm,

$$\frac{\ln x}{\ln(x-12)}=2.$$

Hence,

$$\ln x=2\ln(x-12).$$

Using the property

$$2\ln a=\ln(a^2),$$

we get

$$\ln x=\ln\!\left((x-12)^2\right).$$

Therefore,

$$x=(x-12)^2.$$

Expanding,

$$x=x^2-24x+144.$$
$$x^2-25x+144=0.$$

Factorizing,

$$(x-16)(x-9)=0.$$

So,

$$x=16,\;9.$$

Since

$$x>13,$$

only

$$x=16$$

is valid.

---

✅ Final Answer

$$\boxed{16}$$