JEE Mathematics Logarithms MCQs Set 02

Answer: (b) 1

Explanation:
1. \(\log_2 4=2\)
2. \(\log_2 2=1\)

Teacher's Note:
1. Evaluate the inner logarithm first.
2. The argument of the outer logarithm must be positive, and 2 is.

Answer: (b) -3

Explanation:
1. \(0.001=10^{-3}\)
2. \(\log_{10}10^{-3}=-3\)

Teacher's Note:
1. Write the number as a power of the base.
2. Numbers below 1 have negative logarithms when the base exceeds 1.

Answer: (b) \(-\dfrac52\)

Explanation:
1. \(9\sqrt3=3^{2}\cdot3^{1/2}=3^{5/2}\)
2. Base \(\tfrac13=3^{-1}\)
3. \(\log_{3^{-1}}3^{5/2}=-\dfrac52\)

Teacher's Note:
1. A base below 1 flips the sign when the argument is above 1.
2. Always reduce both to a common prime base.

Answer: (b) 3

Explanation:
1. \(\dfrac18=\left(\dfrac12\right)^3\)
2. So \(\log_{1/2}\dfrac18=3\)

Teacher's Note:
1. Argument and base are on the same side of 1, so the result is positive.
2. Verify by \(\left(\tfrac12\right)^3=\tfrac18\).

Answer: (b) -1

Explanation:
1. \((2+\sqrt3)(2-\sqrt3)=1\), so \(2+\sqrt3=(2-\sqrt3)^{-1}\)
2. \(\log_{2-\sqrt3}(2-\sqrt3)^{-1}=-1\)

Teacher's Note:
1. Numbers that multiply to 1 are reciprocals of each other.
2. The argument is above 1 and the base below 1, so the sign is negative.

Answer: (b) 1

Explanation:
1. Let \(y=\sqrt{5y}\) for the infinite radical
2. \(y^2=5y\), so \(y=5\) (y is positive)
3. \(\log_55=1\)

Teacher's Note:
1. An infinite nested radical satisfies a self-referential equation.
2. Reject the root y = 0 since the radical exceeds 1.

Answer: (c) 4

Explanation:
1. Let the bracket be S, so \(S=\dfrac{1}{3\sqrt2}\sqrt{4-S}\)
2. \(18S^2+S-4=0\), giving \(S=\dfrac{-1\pm17}{36}\); positive root \(S=\dfrac49\)
3. \(\log_{3/2}\dfrac49=-2\)
4. Value \(=6-2=4\)

Teacher's Note:
1. Square the self-similar relation and keep the positive root.
2. \(\tfrac49=\left(\tfrac23\right)^2=\left(\tfrac32\right)^{-2}\).

Answer: (a) 10

Explanation:
1. Antilog gives \(b^{x}\)
2. \(\left(\dfrac1{100}\right)^{-1/2}=100^{1/2}=10\)

Teacher's Note:
1. Antilog base b of x is \(b^x\).
2. A negative exponent inverts the base.

Answer: (b) 6

Explanation:
1. Let \(a=\log_23\). Then \(\log_212=a+2\), \(\log_248=a+4\), \(\log_2192=a+6\)
2. Product \(=(a^2+6a)(a^2+6a+8)\); with \(t=a^2+6a\) the radicand is \(t^2+8t+16=(t+4)^2\)
3. Square root \(=a^2+6a+4\); subtract \((a+2)(a+4)=a^2+6a+8\) to get \(-4\)
4. \(y=-4+10=6\)

Teacher's Note:
1. Express every logarithm through \(\log_23\).
2. The radicand becomes a perfect square, which is the designed trick.

Answer: (a) 1023

Explanation:
1. The product telescopes by base change to \(\log_2(n+1)\)
2. \(\log_2(n+1)=10\), so \(n+1=1024\)
3. \(n=1023\)

Teacher's Note:
1. Change each logarithm to base 2 and cancel.
2. Do not forget to subtract 1 at the end.

Answer: (b) 1

Explanation:
1. Domain requires \(x>3\)
2. \(\log_4(x-1)=\tfrac12\log_2(x-1)\), so \(x-1=(x-3)^2\)
3. \(x^2-7x+10=0\) gives \(x=2\) or \(5\)
4. Only \(x=5\) satisfies \(x>3\), so there is 1 solution

Teacher's Note:
1. Check each root against the domain of every logarithm.
2. The root x = 2 makes \(\log_2(x-3)\) undefined.

Answer: (c) A, B, C and D

Explanation:
1. (A) \(a=2,\ b=1\), so \(x=2\)
2. (B) \(a=\sqrt2,\ b=2\), so \(x=2\)
3. (C) \(a=2,\ b=\log_23\) (irrational), so \(x=3\)
4. (D) \(a=\sqrt2,\ b=2\log_23\) (irrational), so \(x=3\)
5. All four can occur

Teacher's Note:
1. Give an explicit example for each case.
2. Irrational logarithms of integers such as \(\log_23\) are easy to use as exponents.

Answer: (a) true

Explanation:
1. \(\log_35=\dfrac{\ln5}{\ln3}=1.465\) and \(\log_{17}25=\dfrac{\ln25}{\ln17}=1.136\)
2. Since \(1.465>1.136\) the statement is true

Teacher's Note:
1. Compare the values numerically if no neat bound exists.
2. A rough check against 1.5 and 1.1 settles it.

Answer: (a) \(1+\sqrt3\)

Explanation:
1. \((x-1)^2=3\), so \(x-1=\pm\sqrt3\)
2. The base must be positive and not 1, so \(x-1=\sqrt3\)
3. \(x=1+\sqrt3\)

Teacher's Note:
1. A logarithm base must be positive and different from 1.
2. That rejects the negative root.

Answer: (a) \(10^{-5}\) and 1000

Explanation:
1. Take \(\log_{10}\) with \(L=\log x\): \(\dfrac{L(L+5)}{3}=5+L\)
2. \((L+5)\left(\dfrac L3-1\right)=0\)
3. \(L=-5\) or \(L=3\), so \(x=10^{-5}\) or \(1000\)

Teacher's Note:
1. Factor out the common term \(L+5\) instead of cancelling it.
2. Both values are positive, so both are valid.

Answer: (b) 25 and \(\dfrac15\)

Explanation:
1. Take \(\log_5\) with \(t=\log_5x\): \(\dfrac t2(t-1)=1\)
2. \(t^2-t-2=0\), so \(t=2\) or \(t=-1\)
3. \(x=25\) or \(\dfrac15\)

Teacher's Note:
1. \(\log_5\sqrt x=\tfrac12\log_5x\).
2. Both roots are positive and valid.

Answer: (a) 1 and \(\dfrac1{16}\)

Explanation:
1. Let \(L=\log_4x\) and \(y=5^L\): \(5y+\dfrac{1}{5y}=\dfrac{26}{5}\)
2. \(25y^2-26y+1=0\), so \(y=1\) or \(y=\dfrac1{25}\)
3. \(L=0\) or \(-2\), so \(x=1\) or \(\dfrac1{16}\)

Teacher's Note:
1. \(\log_{1/4}x=-\log_4x\).
2. Substitute \(y=5^L\) to turn it into a quadratic.

Answer: (a) 99 and \(-0.9\)

Explanation:
1. Let \(L=\log_{10}(x+1)\): \(L^2=2+L\)
2. \(L=2\) or \(L=-1\)
3. \(x+1=100\) or \(0.1\), so \(x=99\) or \(-0.9\)

Teacher's Note:
1. Take \(\log_{10}\) of both sides to bring down the exponent.
2. Both values satisfy \(x+1>0\).

Answer: (a) \(3+\sqrt6\)

Explanation:
1. Domain: \(x^2>1\), \(x\ne1\), \(x\ne4\)
2. The equation becomes \(\dfrac{x+1}{x-1}=|4-x|\)
3. For \(x<4\): \(x^2-4x+5=0\) has no real root
4. For \(x>4\): \(x^2-6x+3=0\), so \(x=3\pm\sqrt6\); only \(3+\sqrt6\) exceeds 4
5. \(x=3+\sqrt6\)

Teacher's Note:
1. Split the absolute value into two cases and check each root against its case.
2. The root \(3-\sqrt6\) does not belong to the case \(x>4\).

Answer: (a) \(\dfrac13\), 2 and 4

Explanation:
1. Case 1, base equals 1: \(|x-3|=1\) gives \(x=2\) or \(4\)
2. Case 2, exponent zero with non-zero base: \(3x^2-10x+3=0\) gives \(x=3\) or \(\dfrac13\)
3. \(x=3\) makes the base 0, so it is rejected
4. Solutions: \(\dfrac13,\ 2,\ 4\)

Teacher's Note:
1. An expression \(a^b=1\) when \(a=1\), or \(b=0\) with \(a\ne0\) (or \(a=-1\) with even b).
2. Check that the base is not zero in the second case.

Answer: (a) 2 and 81

Explanation:
1. Domain: \(x>0\), \(x\ne1\). For \(02. For \(x>1\): either \(x-1=1\), i.e. \(x=2\), or the exponents are equal
3. With \(t=\log_3x\): \(2t-\dfrac4t=7\), so \(2t^2-7t-4=0\) and \(t=4\) or \(-\dfrac12\); only \(t=4\) (x = 81) has \(x>1\)
4. Solutions: 2 and 81

Teacher's Note:
1. When bases are equal, either the base is 1 or the exponents match.
2. Check the sign of \(x-1\) before removing the absolute value.

Answer: (a) 2, \(2^{-1/3}\) and \(\dfrac14\)

Explanation:
1. Take \(\log_2\) with \(t=\log_2x\): \(t\left(\tfrac34t^2+t-\tfrac54\right)=\tfrac12\)
2. \(3t^3+4t^2-5t-2=0\), which factors as \((t-1)(3t+1)(t+2)=0\)
3. \(t=1,\ -\tfrac13,\ -2\), so \(x=2,\ 2^{-1/3},\ \dfrac14\)

Teacher's Note:
1. Find one rational root of the cubic by trial, then factor.
2. All three values of x are positive, so all are valid.

Answer: (b) 18

Explanation:
1. \(\log5=1-\log2=0.6990\)
2. \(\log5^{25}=25\times0.6990=17.475\)
3. Digits = 17 + 1 = 18

Teacher's Note:
1. Use \(\log5=\log10-\log2\).
2. Characteristic 17 means 18 digits.

Answer: (a) 44

Explanation:
1. \(\alpha_1=4\): \(3^4\le N<3^5\), i.e. \(81\le N<243\)
2. \(\alpha_2=2\): \(25\le N<125\)
3. Both: \(81\le N\le124\), which gives \(124-81+1=44\) integers

Teacher's Note:
1. A characteristic of k means \(b^k\le N2. Intersect the two intervals before counting.

Answer: (a) 342

Explanation:
1. \(3^5\le N<3^6\): \(243\le N<729\)
2. \(5^3\le N<5^4\): \(125\le N<625\)
3. \(7^2\le N<7^3\): \(49\le N<343\)
4. Intersection: \(243\le N<343\), so the largest integer is 342

Teacher's Note:
1. The strict upper bound 343 is excluded.
2. The intersection of intervals uses the largest lower bound and the smallest upper bound.