JEE Mathematics Logarithms MCQs Set 07

Answer: (b) \( 1 \)

Explanation:
1. Each term is \( \log_{abc}\sqrt{bc}=\tfrac12\log_{abc}(bc) \) and so on.
2. Sum \( =\tfrac12\left[\log_{abc}(bc\cdot ca\cdot ab)\right]=\tfrac12\log_{abc}(abc)^2=1 \).

Teacher's Note:
1. Use 1/log_a b = log_b a for each term.
2. The square root halves the logarithm.

Answer: (d) \( \frac{x}{x-1} \)

Explanation:
1. \( 1+\log_ab=x \) so \( \log_ab=x-1 \).
2. \( \log_b(ab)=1+\log_ba=1+\frac1{x-1}=\frac{x}{x-1} \).

Teacher's Note:
1. Reciprocal rule: \( \log_ba=\frac1{\log_ab} \).
2. Split log(ab) as 1 plus a single log.

Answer: (a) \( r^{q/r} \)

Explanation:
1. From the first, \( \log_p(\log_q\log_rx)=0 \) so \( \log_q\log_rx=1 \), \( \log_rx=q \), \( x=r^q \).
2. From the second, \( \log_r\log_px=1 \), \( \log_px=r \), \( x=p^r \).
3. So \( r^q=p^r \) and \( p=r^{q/r} \).

Teacher's Note:
1. Peel the nested logarithms from the outside.
2. Equate the two expressions for x.

Answer: (d) an integer

Explanation:
1. Combine: \( \log_{10}(\log_23\cdot\log_34\cdots\log_{1023}1024) \).
2. The product telescopes: \( \log_23\cdot\log_34\cdots=\log_21024=10 \).
3. So the sum is \( \log_{10}10=1 \), an integer (neither prime nor composite).

Teacher's Note:
1. Use the change of base to telescope products of consecutive logarithms.
2. 1 is neither prime nor composite.

Answer: (d) 200

Explanation:
1. \( x^2=100(2x-75) \) gives \( x^2-200x+7500=0 \), so \( x=50 \) or \( x=150 \).
2. Both satisfy \( 2x-75\gt0 \).
3. Sum \( =200 \).

Teacher's Note:
1. The sum of the roots equals 200 directly from the quadratic.
2. Verify each root against the domain of the logarithm.

Answer: (b) a prime number

Explanation:
1. Let \( t=\log_x5 \): \( \frac t2+t+1=\frac94+\frac{t^2}4 \) gives \( t^2-6t+5=0 \), so \( t=1 \) or \( t=5 \).
2. Then \( x=5 \) or \( x=5^{1/5} \). The positive integer is \( 5 \), a prime.

Teacher's Note:
1. Notice that log_x(5x) = 1 + log_x 5.
2. The second root is not an integer.

Answer: (d) 125

Explanation:
1. \( \sqrt2+\sqrt8=3\sqrt2=\sqrt{18} \), so \( \log_{18}\sqrt{18}=\frac12 \).
2. \( \log_x\frac12=\frac13 \) gives \( x^{1/3}=\frac12 \), so \( x=\frac18 \).
3. Hence \( 1000x=125 \).

Teacher's Note:
1. Simplify the radical first.
2. Convert the final logarithmic equation to exponent form.

Answer: (b) 93

Explanation:
1. \( b=a^2 \) and \( c=b^2=a^4 \).
2. The third condition gives \( \log_3\frac ca=3 \), so \( a^3=27 \) and \( a=3 \).
3. Then \( b=9,\ c=81 \) and \( a+b+c=93 \).

Teacher's Note:
1. Express b and c in terms of a.
2. Combine the log terms on one side.

Answer: (b) \( -\frac53 \)

Explanation:
1. \( x=\frac{-\ln5}{\ln3}\cdot\frac{3\ln7}{3\ln5}\cdot\frac{6\ln3}{2\ln7}=-3 \).
2. The exponent of 25: \( 3\cdot\frac{\ln11}{2\ln17}\cdot\frac{\ln17}{2\ln28}\cdot\frac{\ln784}{3\ln11}=\frac14\cdot\frac{\ln784}{\ln28}=\frac12 \), so \( y=25^{1/2}=5 \).
3. Hence \( \frac yx=-\frac53 \).

Teacher's Note:
1. Write every logarithm as a ratio of natural logarithms and cancel.
2. Note 784 = 28 squared.

Answer: (c) \( -\frac14 \)

Explanation:
1. \( 6x^2+23x+21=(2x+3)(3x+7) \) and \( 4x^2+12x+9=(2x+3)^2 \). Let \( u=\log_{2x+3}(3x+7) \).
2. The equation is \( 1+u=4-\frac2u \), i.e. \( u^2-3u+2=0 \), so \( u=1 \) or \( 2 \).
3. If \( u=1 \): \( x=-4 \) (rejected, base negative). If \( u=2 \): \( 4x^2+9x+2=0 \) gives \( x=-\frac14 \) or \( -2 \); \( x=-2 \) is rejected, so \( x=-\frac14 \).

Teacher's Note:
1. Always check bases are positive and not equal to 1.
2. Factor the quadratic arguments first.

Answer: (d) \( \frac{\alpha\beta}{\beta-\alpha} \)

Explanation:
1. \( \log_{p/q}x=\frac1{\log_x(p/q)}=\frac1{\log_xp-\log_xq} \).
2. \( \log_xp=\frac1\alpha,\ \log_xq=\frac1\beta \).
3. So the value is \( \dfrac1{\frac1\alpha-\frac1\beta}=\dfrac{\alpha\beta}{\beta-\alpha} \).

Teacher's Note:
1. Flip to base x using the reciprocal rule.
2. Take care with the sign of the difference.

Answer: (b) \( \log_25 \)

Explanation:
1. \( (2^x-1)^2=2(2^x+3) \).
2. With \( t=2^x \): \( t^2-4t-5=0 \), so \( t=5 \) (as \( t\gt1 \)).
3. Hence \( x=\log_25 \).

Teacher's Note:
1. In an A.P. the middle term squared (inside logs, the product) matches the outer terms.
2. Reject the negative root of t.

Answer: (c) 2, 3

Explanation:
1. Check the options: \( x=2 \): \( \log_21=0=2-2 \) holds.
2. For \( x=3 \): \( \log_22=1=3-2 \) holds.
3. For \( x=4 \): \( \log_25\ne2 \). So the roots are 2 and 3.

Teacher's Note:
1. The equation x² - 4x + 5 = 2^(x-2) has two intersection points.
2. Substitution of the options is the quickest check.

Answer: (c) \( \left[\frac12,1\right)\cup\left(2,\frac52\right] \)

Explanation:
1. Base \( \sin\frac\pi3=\frac{\sqrt3}2\lt1 \), so the inequality reverses: \( 0\lt x^2-3x+2\le\frac34 \).
2. The left part gives \( x\lt1 \) or \( x\gt2 \); the right part \( x^2-3x+\frac54\le0 \) gives \( \frac12\le x\le\frac52 \).
3. Solution: \( \left[\frac12,1\right)\cup\left(2,\frac52\right] \).

Teacher's Note:
1. For a base between 0 and 1 reverse the inequality.
2. Always keep the condition that the argument is positive.

Answer: (a) \( (2,\infty) \)

Explanation:
1. \( \log_{0.09}(x-1)=\frac12\log_{0.3}(x-1) \). Let \( L=\log_{0.3}(x-1) \).
2. Then \( L\lt\frac L2 \) gives \( L\lt0 \).
3. Since the base is below 1, \( x-1\gt1 \), i.e. \( x\gt2 \).

Teacher's Note:
1. Bring both logarithms to the same base 0.3.
2. A negative log to a base less than 1 means the argument exceeds 1.

Answer: (a) \( (-3,-\sqrt5)\cup(\sqrt5,3) \)

Explanation:
1. \( \log_{0.5}1=0 \), so with base below 1 we need \( 0\lt\log_5(x^2-4)\lt1 \).
2. This gives \( 1\lt x^2-4\lt5 \), i.e. \( 5\lt x^2\lt9 \).
3. So \( \sqrt5\lt|x|\lt3 \).

Teacher's Note:
1. Both the inner and the outer logarithm impose a condition.
2. |x| between two positive numbers gives two intervals.

Answer: (d) \( (3,\infty) \)

Explanation:
1. Base is below 1: \( x^2-2x\gt2 \), i.e. \( x^2-2x-2\gt0 \).
2. So \( x\lt1-\sqrt3\approx-0.73 \) or \( x\gt1+\sqrt3\approx2.73 \).
3. The set \( (3,\infty) \) lies inside the solution set.

Teacher's Note:
1. Exponential with base below 1 reverses the inequality.
2. Approximate the surd to compare with the given intervals.

Answer: (d) \( \varphi \)

Explanation:
1. The base requires \( 1-x\gt0 \) and \( 1-x\ne1 \), so \( x\lt1 \).
2. The argument requires \( x-2\gt0 \), so \( x\gt2 \).
3. These cannot hold together, so the solution set is empty.

Teacher's Note:
1. Check the domain before solving.
2. A contradictory domain means no solution.

Answer: (d) 1

Explanation:
1. Let \( x=\ln a,\ y=\ln b,\ z=\ln c \). The left side is \( \frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz} \).
2. So \( x^3+y^3+z^3-3xyz=0 \), i.e. \( (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0 \).
3. As the numbers are distinct the second factor is non-zero, so \( x+y+z=0 \) and \( abc=1 \).

Teacher's Note:
1. Use the identity for the sum of cubes with a zero value.
2. Distinct numbers rule out the second factor.

Answer: (c) 3.83

Explanation:
1. \( b=a^{3/2} \) and \( d=c^{5/4} \), so \( a=m^2,\ b=m^3,\ c=n^4,\ d=n^5 \) for integers \( m,n \).
2. Then \( m^2-n^4=9 \) gives \( (m-n^2)(m+n^2)=9 \), so \( m-n^2=1,\ m+n^2=9 \), i.e. \( m=5,\ n=2 \).
3. Thus \( a=25,b=125,c=16,d=32 \) and the value is \( \frac{157}{41}\approx3.83 \).

Teacher's Note:
1. Integer powers: a must be a perfect square, c a perfect fourth power.
2. Factorise the difference of squares.

Answer: (a) 12.50

Explanation:
1. \( 2x^2-21x+50=100 \) gives \( 2x^2-21x-50=0 \).
2. The roots are \( x=\frac{21\pm29}4=12.5 \) or \( -2 \).
3. The positive root is \( 12.5 \).

Teacher's Note:
1. Convert the logarithm to exponent form.
2. The discriminant is the perfect square 841.

Answer: (c) 0.25

Explanation:
1. Let \( u=\log_{10}x \): \( u^2+2u+1=\log_{10}^22 \), so \( (u+1)^2=(\log_{10}2)^2 \).
2. Hence \( u=-1\pm\log_{10}2 \), i.e. \( x=\frac{2}{10}=0.2 \) or \( x=\frac1{20}=0.05 \).
3. Sum \( =0.25 \).

Teacher's Note:
1. Complete the square in u.
2. Convert back from u to x with powers of 10.

Answer: (d) 1

Explanation:
1. Take \( \log_{2009} \) of both sides. Let \( u=\log_{2009}x \) and \( k=\log_{2009}2010 \): \( 1+u-k=\frac ku \).
2. So \( u^2+(1-k)u-k=0 \), i.e. \( (u+1)(u-k)=0 \), giving \( x=\frac1{2009} \) or \( x=2010 \).
3. Product \( =\frac{2010}{2009} \), so \( m-n=1 \).

Teacher's Note:
1. Take logarithms to base 2009.
2. Factorise the quadratic in u.

Answer: (b) 3

Explanation:
1. Take \( \log_2 \) and put \( u=\log_2x \): \( u\left(\frac34u^2+u-\frac54\right)=\frac12 \), i.e. \( 3u^3+4u^2-5u-2=0 \).
2. This factorises as \( (u-1)(3u+1)(u+2)=0 \), so \( u=1,-\frac13,-2 \).
3. Hence \( x=2,\ 2^{-1/3},\ \frac14 \): three real solutions, exactly one of which is irrational.

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

Answer: (c) an irrational number

Explanation:
1. Suppose \( \log_27=\frac pq \) with integers \( p,q \). Then \( 2^p=7^q \).
2. The left side is even and the right side is odd (for \( p,q\gt0 \)), a contradiction.
3. So \( \log_27 \) is irrational.

Teacher's Note:
1. Reduce to the equality of an even and an odd number.
2. This is a standard proof for logs of coprime integers.

Answer: (d) 3

Explanation:
1. Note \( \log_{1/4}(x^2+7)^{-1}=\log_4(x^2+7) \). Try \( x^2+7=16 \).
2. Then \( \log_816=\frac43,\ \log_{3/4}\frac43=-1 \) and \( \log_416=2,\ \log_{1/2}2=-1 \).
3. The sum is \( -2 \) as required, so \( x^2=9 \) and \( x=3 \) (the other root is \( -3 \)).

Teacher's Note:
1. The left side is increasing in x², so there is only one value of x² that works.
2. Use the substitution x² + 7 = 16 as a trial.

Answer: (b) 1

Explanation:
1. \( \log_4(x-1)=\frac12\log_2(x-1) \), so \( x-1=(x-3)^2 \).
2. This gives \( x^2-7x+10=0 \), so \( x=2 \) or \( 5 \).
3. Domain: \( x\gt3 \), so only \( x=5 \): exactly one solution.

Teacher's Note:
1. Convert to a common base.
2. Reject roots outside the domain of the logarithms.