JEE Mathematics Logarithms MCQs Set 08

Answer: (c) 4

Explanation:
1. Let the nested expression inside the brackets be \( S \). Then \( S=\frac1{3\sqrt2}\sqrt{4-S} \).
2. So \( 18S^2=4-S \), i.e. \( 18S^2+S-4=0 \), giving \( S=\frac{-1+17}{36}=\frac49 \).
3. Then \( \log_{3/2}\frac49=-2 \) and the value is \( 6-2=4 \).

Teacher's Note:
1. The pattern repeats, so the whole expression satisfies a self-referential equation.
2. Reject the negative root of the quadratic.

Answer: (d) \( \frac{2\log_23}{2\log_23-1} \)

Explanation:
1. Taking logs, \( x\ln3=(x-1)\ln4 \), so \( x=\frac{\ln4}{\ln4-\ln3}\approx4.82 \).
2. Dividing by \( \ln3 \) gives (a); dividing by \( \ln4 \) gives (c); dividing by \( \ln2 \) gives (b).
3. Option (d) equals \( \frac{2\log_23}{2\log_23-1}\approx1.46 \), so it is not equal to \( x \).

Teacher's Note:
1. Test each expression after converting logs to natural logarithms.
2. Changing the base of the whole fraction by one common divisor keeps its value.

Answer: (a) 2

Explanation:
1. Let \( t=\log_2x \): \( \frac2t+\frac6{1+t}=3 \).
2. This gives \( 3t^2-5t-2=0 \), so \( t=2 \) or \( t=-\frac13 \).
3. Hence \( x=4 \) or \( x=2^{-1/3} \): two real solutions (one integral, one irrational).

Teacher's Note:
1. Change both logs to base 2.
2. Check that bases are positive and not 1.

Answer: (c) \( 1,\ 5,\ \frac1{25} \)

Explanation:
1. Let \( u=\log_5x \): \( u^2+\frac{1-u}{1+u}=1 \).
2. This gives \( u^3+u^2-2u=0 \), i.e. \( u(u-1)(u+2)=0 \).
3. So \( x=5^0,5^1,5^{-2} \): \( 1,\ 5,\ \frac1{25} \).

Teacher's Note:
1. Write log_{5x}(5/x) as (1 - u)/(1 + u).
2. Check that 5x is a valid base: it is positive and not 1 for these values.

Answer: (b) (2) and (3) only

Explanation:
1. (1): \( \log_23\approx1.585 \) and \( \log_{12}10\approx0.927 \): false.
2. (2): \( \log_65\approx0.898\lt1 \) and \( \log_78\approx1.069\gt1 \): true. (3): \( \log_326\approx2.97\lt\log_29\approx3.17 \): true.
3. (4): \( \log_{16}15\approx0.977\lt1 \) while \( \log_{10}11\approx1.041\gt1 \): false.

Teacher's Note:
1. Compare each logarithm with 1 or with an integer.
2. Use a calculator-free estimate: log_b a is below 1 when a < b.

Answer: (a) (1), (2) and (4) only

Explanation:
1. Base \( 0.1\lt1 \), so the inequality reverses: \( (0.1)^2\le x\le(0.1)^{1/2} \).
2. So \( \frac1{100}\le x\le\frac1{\sqrt{10}} \).
3. Maximum \( \frac1{\sqrt{10}} \), minimum \( \frac1{100} \): statements (1), (2) and (4) are correct.

Teacher's Note:
1. With a base below 1 larger log means smaller x.
2. Convert both bounds to powers of 0.1.

Answer: (c) In all four cases

Explanation:
1. \( x=a^b \). Case (1): \( a=2^{\sqrt2},\ b=\frac1{\sqrt2} \) gives \( x=2 \). Case (2): \( a=2,\ b=\log_23 \) gives \( x=3 \).
2. Case (3): \( a=\sqrt2,\ b=2 \) gives \( x=2 \). Case (4): \( a=2,\ b=3 \) gives \( x=8 \).
3. So x can be rational in every case.

Teacher's Note:
1. Choose the exponent so that it cancels the irrationality of the base.
2. An irrational power of an irrational number can be rational.

Answer: (c) 3

Explanation:
1. Take \( \log_3 \) and let \( u=\log_3x \): \( u\left(u^2-\frac92u+5\right)=\frac32 \), i.e. \( 2u^3-9u^2+10u-3=0 \).
2. This factorises as \( (u-1)(2u-1)(u-3)=0 \), so \( u=1,\frac12,3 \).
3. So \( x=3,\ \sqrt3,\ 27 \): three positive solutions.

Teacher's Note:
1. Test u = 1 to find the first factor of the cubic.
2. Each u gives exactly one x.

Answer: (a) 6

Explanation:
1. For A: put \( t=\log_4x \): \( 1+t+4(5-4t)=3(5-4t)(1+t) \) gives \( 2t^2-3t+1=0 \), so \( t=1,\frac12 \), \( x=4,2 \) and \( A=6 \).
2. For B: \( m=\log_23,\ n=\log_34 \), so \( mn=\log_24=2 \).
3. For C: put \( u=\log_3x \): \( \frac{1-u}{1+u}+u^2=1 \) gives \( u(u-1)(u+2)=0 \), so \( x=1,3,\frac19 \); integral roots 1 and 3, so \( C=4 \).
4. Hence \( B+C=2+4=6 \).

Teacher's Note:
1. The integral roots are 1 and 3; the root 1/9 is not an integer.
2. B uses the change of base formula.

Answer: (b) 8

Explanation:
1. For A: put \( t=\log_4x \): \( 1+t+4(5-4t)=3(5-4t)(1+t) \) gives \( 2t^2-3t+1=0 \), so \( t=1,\frac12 \), \( x=4,2 \) and \( A=6 \).
2. For B: \( m=\log_23,\ n=\log_34 \), so \( mn=\log_24=2 \).
3. For C: put \( u=\log_3x \): \( \frac{1-u}{1+u}+u^2=1 \) gives \( u(u-1)(u+2)=0 \), so \( x=1,3,\frac19 \); integral roots 1 and 3, so \( C=4 \).
4. Hence \( A+C\div B=6+\frac42=8 \).

Teacher's Note:
1. Mind the order of operations: divide before adding.
2. Use the values A = 6, B = 2, C = 4.

Answer: (b) 1

Explanation:
1. For \( x\gt0 \), \( y=\log_{1/2}x \) is decreasing from \( +\infty \) to \( -\infty \) while \( y=7^x \) is increasing and always above 1.
2. At \( x\to0^+ \) the log curve is above the exponential curve and at \( x=1 \) it is below (\( 0\lt7 \)).
3. So the graphs meet exactly once.

Teacher's Note:
1. One decreasing and one increasing function intersect at most once.
2. Check the two ends of the interval to confirm they cross.

Answer: (a) 2

Explanation:
1. \( \sqrt{8\pm2\sqrt7}=\sqrt7\pm1 \), so \( a=3\cdot2=6 \).
2. Also \( b=\sqrt{1260+36}=\sqrt{1296}=36 \).
3. So \( \log_636=2 \).

Teacher's Note:
1. Write 8 ± 2√7 as a perfect square.
2. 1296 = 36 squared.

Answer: (c) -1

Explanation:
1. \( 3\pm2\sqrt2=(\sqrt2\pm1)^2 \), so \( a=\sqrt2+1 \) and \( b=\sqrt2-1 \).
2. Since \( ab=1 \), \( b=a^{-1} \), hence \( \log_ab=-1 \).

Teacher's Note:
1. If two positive numbers have product 1 then each is the reciprocal of the other.
2. Recognise the perfect squares under the radicals.

Answer: (b) -1

Explanation:
1. \( ab=\sqrt{7^2-(7^2-1)}=1 \).
2. So \( b=\frac1a \) and \( \log_ab=-1 \).

Teacher's Note:
1. Multiply the two radicals first.
2. A product of 1 means the numbers are reciprocals.

Answer: (c) 1

Explanation:
1. \( (2^x-3^y)(2^x+3^y)=55 \) with positive factors, so the factor pairs are \( 1\times55 \) and \( 5\times11 \).
2. For \( 1,55 \): \( 2^x=28 \), impossible. For \( 5,11 \): \( 2^x=8,\ 3^y=3 \), so \( x=3,\ y=1 \).
3. Exactly one solution.

Teacher's Note:
1. Factorise the difference of squares.
2. Powers of 2 and 3 must match the factor sums.

Answer: (d) 15

Explanation:
1. \( 0.\overline{36}=\frac{36}{99}=\frac4{11} \).
2. Sum \( 4+11=15 \).

Teacher's Note:
1. A two-digit repeating block over 99 gives the fraction.
2. Reduce to the lowest terms before adding.

Answer: (c) 16

Explanation:
1. Let \( x=\log_8a,\ y=\log_8b \). Then \( \log_ab=\frac yx=3 \), so \( y=3x \).
2. The first condition gives \( 4x=3x^2 \), so \( x=\frac43 \) (since \( x\ne0 \)).
3. Thus \( a=8^{4/3}=16 \).

Teacher's Note:
1. Express everything through x = log_8 a.
2. Reject x = 0 because then a = 1 and log_a b is undefined.

Answer: (a) 0

Explanation:
1. Let \( s=\sqrt{\log_32} \), so \( \sqrt{\log_23}=\frac1s \).
2. Then \( 2^{1/s}=3^{\log_32/s}=3^{s^2/s}=3^s \).
3. Hence \( P=3^s-3^s=0 \).

Teacher's Note:
1. Write 2 as 3^(log_3 2).
2. The two terms are equal.

Answer: (a) 10

Explanation:
1. \( 2^{10}=1024\lt2008\lt2048=2^{11} \).
2. So \( 10\lt\log_22008\lt11 \) and the integral part (characteristic) is 10.

Teacher's Note:
1. Locate the number between consecutive powers of the base.
2. The characteristic is the integral part of the logarithm.

Answer: (a) 5

Explanation:
1. \( \log_{10}16\cdot\ln10=\ln16 \).
2. So \( \ln2\cdot\frac{\ln625}{\ln b}=\ln16 \), giving \( \ln b=\frac{\ln2\cdot4\ln5}{4\ln2}=\ln5 \).
3. Hence \( b=5 \).

Teacher's Note:
1. Use log_10 16 times ln 10 equals ln 16.
2. 625 = 5^4 and 16 = 2^4.

Answer: (d) 1

Explanation:
1. \( 75600=2^4\cdot3^3\cdot5^2\cdot7 \).
2. The sum is \( \log_n2^4+\log_n3^3+\log_n5^2+\log_n7=\log_n75600=1 \).

Teacher's Note:
1. Prime-factorise n first.
2. Reciprocal of a log is the log with swapped base and argument.

Answer: (b) 89

Explanation:
1. First: \( \log_3\log_4x=1 \), \( \log_4x=3 \), so \( x=64 \).
2. Second: \( \log_4\log_2y=1 \), \( \log_2y=4 \), so \( y=16 \).
3. Third: \( \log_2\log_3z=1 \), \( \log_3z=2 \), so \( z=9 \). Sum \( =89 \).

Teacher's Note:
1. Work from the outermost logarithm inwards.
2. Each 'log = 0' step gives the argument 1.

Answer: (a) 16

Explanation:
1. Let \( t=\log_2x \): \( \log_4t=\frac12\log_2t \) and \( \log_4x=\frac t2 \).
2. The equation is \( \frac12\log_2t+\log_2\frac t2=2 \), i.e. \( \frac32\log_2t-1=2 \), so \( \log_2t=2 \) and \( t=4 \).
3. Hence \( x=2^4=16 \).

Teacher's Note:
1. Reduce all logs to base 2.
2. Check: log_2 16 = 4 and log_4 16 = 2.

Answer: (d) \( \frac13 \)

Explanation:
1. Convert to exponent form: \( \log_9x+\frac12+9^x=3^{2x}=9^x \).
2. So \( \log_9x=-\frac12 \), which gives \( x=9^{-1/2}=\frac13 \).

Teacher's Note:
1. Note that 3^(2x) equals 9^x, so the exponential terms cancel.
2. Convert the remaining log equation to exponent form.

Answer: (d) 0

Explanation:
1. \( \log_{\sqrt x}(2x)=2\log_x(2x) \), so \( x^{2\log_x(2x)}=(2x)^2=4 \).
2. This gives \( x=\pm1 \).
3. But \( x=1 \) is not a valid base and \( x=-1 \) is negative, so there is no real root.

Teacher's Note:
1. Check the base conditions: x > 0 and x ≠ 1.
2. The algebra gives candidates only.

Answer: (d) 2

Explanation:
1. \( x^{0.5\cdot2\log_x(x^2-x)}=x^{\log_x(x^2-x)}=x^2-x \).
2. The right side is \( 3^{\log_32}=2 \).
3. So \( x^2-x-2=0 \) gives \( x=2 \) or \( -1 \); the base must be positive, so \( x=2 \).

Teacher's Note:
1. Use a^(log_a b) = b.
2. log_9 4 = log_3 2.

Answer: (a) \( 0\lt f(x)\lt1 \)

Explanation:
1. All four factors of \( \frac{(x-1)(x-5)}{(x-2)(x-3)} \) are positive for \( x\gt5 \), so \( f(x)\gt0 \).
2. Also \( f(x)-1=\frac{-x-1}{(x-2)(x-3)}\lt0 \), so \( f(x)\lt1 \).
3. Hence \( 0\lt f(x)\lt1 \).

Teacher's Note:
1. Compare f(x) with 1 by subtracting 1 and simplifying.
2. The numerator of f(x) - 1 simplifies to -x - 1.