JEE Maths DPP
Change of Base and Evaluation
DPP: LOG-CHBASE-2026-001
Part A: Multiple Choice Questions (Q1–Q13)
-
The value of $\log_{3} 5 \cdot \log_{25} 27$ is:
- $3/2$
- $2/3$
- $1$
- $9/4$
-
If $a = \log_{24} 48$ and $b = \log_{36} 72$, then the value of $2ab – (a+b)$ is:
- $1$
- $0$
- $\log_{24} 36$
- None of these
-
If $\log_{a} b = 10$, then the value of $\log_{b^5} a^2$ is:
- $4$
- $25$
- $1/25$
- $0.04$
-
The expression $\frac{1}{\log_{3} 2} + \frac{1}{\log_{4} 2} + \dots + \frac{1}{\log_{10} 2}$ simplifies to:
- $\log_{2} (10!)$
- $\log_{2} (10!/2)$
- $\log_{2} (5!)$
- $10!$
-
If $x = \log_{c} (ab)$, $y = \log_{a} (bc)$, and $z = \log_{b} (ca)$, then
$$
\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1}
$$
is:
- $1$
- $2$
- $0$
- $3$
-
The value of $\log_{4} 18$ is:
- $1 + 2\log_{4} 3$
- $\frac{1}{2} + \log_{2} 3$
- $\frac{1}{2} + \log_{4} 9$
- Both (b) and (c)
-
If $\log_{12} 27 = a$, then $\log_{6} 16$ in terms of $a$ is:
- $\frac{4(3-a)}{3+a}$
- $\frac{4(3+a)}{3-a}$
- $\frac{3-a}{3+a}$
- $\frac{a}{3-a}$
-
Find the value of $\log_{5} \sqrt{5 \sqrt{5 \sqrt{5 \dots \infty}}}$:
- $1/2$
- $1$
- $0$
- $\infty$
-
If $\log_{3} 2, \log_{3} (2^x – 5), \log_{3} (2^x – 7/2)$ are in A.P., then $x$ is:
- $2$
- $3$
- $4$
- $5$
-
The value of $81^{1/\log_{5} 3} + 27^{\log_{9} 36} + 3^{4/\log_{7} 9}$ is:
- $890$
- $930$
- $720$
- $1024$
-
If $\log_{10} 2 = 0.3010$, then the number of digits in $5^{20}$ is:
- $14$
- $15$
- $13$
- $16$
-
Let $S = \sum_{n=2}^{100} \frac{1}{\log_{n} 100!}$. The value of $S$ is:
- $0$
- $1$
- $100$
- $1/2$
-
If $x = \log_{2} 3 \cdot \log_{3} 4 \cdot \log_{4} 5 \dots \log_{n} (n+1) = 10$, then $n$ is:
- $1023$
- $1024$
- $511$
- $2047$
Part B: Subjective (Q14–Q15)
- Solve the equation: $$ 4^{\log_{9} x} – 6 \cdot 2^{\log_{9} x} + 8 = 0. $$
- Prove that $$ \frac{\log_{a} n}{\log_{ab} n} = 1 + \log_{a} b $$ and hence find the value of $\log_{3} 10$ if $\log_{30} 10 = a$.
Part C: Integer/Assertion Reason
-
Assertion: $\log_{2} 3$ is irrational.
Reason: If $\log_{a} b$ is rational, say $p/q$, then $a^p = b^q$. For $a=2, b=3$, $2^p = 3^q$ has no integer solution for $p, q > 0$.- A
- B
- C
- D
- Integer: Find the value of $x$ satisfying $$ \log_{x} 2 \cdot \log_{x/16} 2 = \log_{x/64} 2. $$
- Integer: If $\log_{2} (\log_{2} (\log_{3} x)) = \log_{2} (\log_{3} (\log_{2} y)) = 0$, find $x-y$.
- Integer: Find the value of $$ 7^{\log_{7} 11} + \log_{2} 8. $$
- Integer: If $a^2 + b^2 = 7ab$, find the value of $k$ if $$ 2\log\left(\frac{a+b}{3}\right) = k(\log a + \log b). $$
-
Assertion: $\log_{10} 2$ lies between $1/4$ and $1/3$.
Reason: $2^3 < 10 < 2^4$. - Integer: Calculate $$ \log_{3} 4 \cdot \log_{4} 5 \cdot \log_{5} 6 \cdot \log_{6} 7 \cdot \log_{7} 8 \cdot \log_{8} 9. $$
Answer Key
| Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 | Q11 |
|---|---|---|---|---|---|---|---|---|---|---|
| A | B | D | B | A | D | A | B | B | B | A |
| Q12 | Q13 | Q14 | Q15 | Q16 | Q17 | Q18 | Q19 | Q20 | Q21 | Q22 |
| B | A | 81, 16 | $\frac{a}{1-a}$ | A | 4 | 1 | 14 | 1 | A | 2 |