JEE Mathematics Logarithms MCQs Set 03

Answer: (b) \(32\)

Explanation:
1. \(\log_{\sqrt8}b=\dfrac{10}{3}\) and \(\log_{\sqrt8}b=\dfrac{\log_2b}{\log_2 2^{3/2}}=\dfrac23\log_2b\)
2. So \(\log_2b=\dfrac{10}{3}\cdot\dfrac32=5\)
3. \(b=2^5=32\)

Teacher's Note:
1. Convert the base \(\sqrt8=2^{3/2}\) to base 2 first.
2. Check: \(\sqrt8^{10/3}=2^{5}=32\).

Answer: (a) \(1\)

Explanation:
1. \(1+\log_abc=\log_aa+\log_abc=\log_a(abc)\), so the first term is \(\dfrac1{\log_a(abc)}=\log_{abc}a\)
2. In the same way the other two terms are \(\log_{abc}b\) and \(\log_{abc}c\)
3. Sum \(=\log_{abc}(abc)=1\)

Teacher's Note:
1. Write 1 as \(\log_aa\) so that the denominator becomes a single logarithm.
2. Then use the reciprocal rule \(1/\log_pq=\log_qp\).

Answer: (c) \(2a-3b-1\)

Explanation:
1. \(b-a=\log\dfrac{225}{25}=\log9\), so \(2\log3=b-a\)
2. The expression is \(-2\log9-\log2250=-2(b-a)-\log(225\times10)=-2(b-a)-(b+1)\)
3. \(=-2b+2a-b-1=2a-3b-1\)

Teacher's Note:
1. Express 9 and 2250 through 25 and 225 so only a and b appear.
2. The constant \(\log10=1\) is what gives the final \(-1\).

Answer: (b) \(\dfrac12\)

Explanation:
1. \(\dfrac1{\log_236}=\log_{36}2\) and \(\dfrac1{\log_336}=\log_{36}3\)
2. Sum \(=\log_{36}6=\dfrac12\), since \(36^{1/2}=6\)

Teacher's Note:
1. Reciprocal of a logarithm swaps its base and argument.
2. Then combine with the product rule.

Answer: (a) \((3,4)\cup(4,\infty)\)

Explanation:
1. Base: \(x-3>0\) and \(x-3\ne1\), so \(x>3\), \(x\ne4\)
2. Argument: \(2x-3>0\), so \(x>\dfrac32\)
3. Combining: \(x\in(3,4)\cup(4,\infty)\)

Teacher's Note:
1. Three conditions: positive base, base not equal to 1, positive argument.
2. Take the intersection of all three.

Answer: (d) \(\dfrac{4(3-a)}{3+a}\)

Explanation:
1. \(a=\dfrac{3\log_23}{2+\log_23}\), so \(\log_23=\dfrac{2a}{3-a}\)
2. \(\log_616=\dfrac{4}{\log_26}=\dfrac4{1+\log_23}\)
3. \(=\dfrac{4}{1+\frac{2a}{3-a}}=\dfrac{4(3-a)}{3+a}\)

Teacher's Note:
1. Everything reduces to \(\log_23\), so find it from the given condition first.
2. Note that a cannot equal 3, otherwise \(\log_23\) would be undefined.

Answer: (c) characteristic \(-17\), mantissa \(0.52\)

Explanation:
1. \(\log x=20\log\dfrac{15}{100}=20(\log3+\log5-2)\), and \(\log5=1-\log2\)
2. \(=20(0.477+0.699-2)=20(-0.824)=-16.48\)
3. Write \(-16.48=-17+0.52\): the mantissa must lie in \([0,1)\), so characteristic \(-17\), mantissa \(0.52\)

Teacher's Note:
1. The mantissa is never negative, so round the characteristic down.
2. A common slip is to write the characteristic as -16 and the mantissa as -0.48.

Answer: (b) \(2058\)

Explanation:
1. Characteristic 3 means \(3\le\log_7N<4\)
2. So \(7^3\le N<7^4\), i.e. \(343\le N\le2400\)
3. Count \(=2401-343=2058\)

Teacher's Note:
1. For characteristic k in base b, the count is \(b^{k+1}-b^k\).
2. Do not include \(7^4\) itself because its characteristic is 4.

Answer: (d) \(x=-3\)

Explanation:
1. Equate the arguments: \(x^2-3x-5=7-2x\), so \(x^2-x-12=0\) and \(x=4\) or \(-3\)
2. Domain needs \(7-2x>0\), i.e. \(x<3.5\)
3. \(x=4\) gives \(7-8=-1<0\) and is rejected; \(x=-3\) gives \(13>0\) on both sides. So \(x=-3\)

Teacher's Note:
1. After removing the logarithms always test each root in the original equation.
2. Here one root violates positivity of the argument.

Answer: (c) \(0<x<1\)

Explanation:
1. Since \(a\ne0\), \(a^2+1>1\)
2. The logarithm of a number greater than 1 is negative only when the base lies between 0 and 1
3. So \(0<x<1\)

Teacher's Note:
1. Positive and negative logarithms depend on which side of 1 the base and number lie.
2. Opposite sides of 1 give a negative logarithm.

Answer: (b) \((0,1)\cup(2,\infty)\)

Explanation:
1. Put \(t=\log_2x\). The inequality becomes \(\dfrac{-1}{t(t-1)}<1\), i.e. \(\dfrac{t^2-t+1}{t(t-1)}>0\)
2. \(t^2-t+1=\left(t-\dfrac12\right)^2+\dfrac34>0\), so we need \(t(t-1)>0\)
3. \(t<0\) or \(t>1\), i.e. \(0<x<1\) or \(x>2\)

Teacher's Note:
1. Substitute \(t=\log_2x\) to turn the inequality into a rational one in t.
2. Exclude x = 1 and x = 2 where a denominator vanishes.

Answer: (a) \(6\)

Explanation:
1. The expression equals \(\log_z[(x^2-y^2)(x^2+y^2-xy)(x^2+y^2+xy)]\)
2. \((x^2-y^2)=(x-y)(x+y)\), and \((x+y)(x^2-xy+y^2)=x^3+y^3\), \((x-y)(x^2+xy+y^2)=x^3-y^3\)
3. Product \(=x^6-y^6=z^6\), so the value is \(\log_zz^6=6\)

Teacher's Note:
1. Factorise using the sum and difference of cubes.
2. Combine the logarithms into one before simplifying.

Answer: (d) \(-6\)

Explanation:
1. \(N=1.65\times10^{-6}\)
2. \(\log N=\log1.65+(-6)\) with \(0\le\log1.65<1\)
3. The characteristic is \(-6\) (mantissa \(\log1.65\))

Teacher's Note:
1. Write the number in standard form \(m\times10^k\); the characteristic is k.
2. Count the zeros after the decimal point and add 1 to get the magnitude of k.

Answer: (a) \(x=0\)

Explanation:
1. \(\log_2(9-2^x)=3-x\), so \(9-2^x=2^{3-x}=\dfrac8{2^x}\)
2. Put \(t=2^x\): \(t^2-9t+8=0\), so \(t=1\) or \(8\), i.e. \(x=0\) or \(3\)
3. \(x=3\) makes the denominator \(3-x\) zero, so only \(x=0\) is valid

Teacher's Note:
1. The denominator is zero at x = 3, which removes that root.
2. Always test each root against the domain of the original equation.

Answer: (d) \(x=-5\)

Explanation:
1. \(x^2-2x+65=(5-x)^2=x^2-10x+25\)
2. \(8x=-40\), so \(x=-5\)
3. Base \(5-x=10>0\) and \(\ne1\), argument \(100>0\): valid

Teacher's Note:
1. A logarithmic equation with a variable base gives argument = base squared.
2. Always test each root against the domain of the original equation.

Answer: (c) \(x=\dfrac13\)

Explanation:
1. \(2x=\log_39^x\), so \(\log_9x+\dfrac12+9^x=9^x\)
2. \(\log_9x=-\dfrac12\), so \(x=9^{-1/2}=\dfrac13\)

Teacher's Note:
1. Notice that 9^x cancels from both sides.
2. Then a single logarithm remains.

Answer: (b) \(x=0\)

Explanation:
1. \((x+1)(x+3)=3\), so \(x^2+4x=0\) and \(x=0\) or \(-4\)
2. Domain needs \(x>-1\), so \(x=-4\) is rejected; \(x=0\)

Teacher's Note:
1. Always test each root against the domain of the original equation.
2. Combine two logarithms of the same base into a product.

Answer: (a) \(x=3\)

Explanation:
1. Put \(t=2^x\): \((t-1)(t-7)=7\), so \(t^2-8t=0\) and \(t=0\) or \(8\)
2. Domain needs \(t>7\), so \(t=8\) and \(x=3\)

Teacher's Note:
1. Substitute t = 2^x and then convert back.
2. Always test each root against the domain of the original equation.

Answer: (c) \(x=\dfrac{16}{\sqrt[3]5}\)

Explanation:
1. \(1-\log5=\log2\), so \(3\log2=\log\dfrac{x\cdot5^{1/3}}{2}\)
2. \(8=\dfrac{x\cdot5^{1/3}}2\)
3. \(x=\dfrac{16}{5^{1/3}}\)

Teacher's Note:
1. Write 1 as log 10 so 1 - log 5 becomes log 2.
2. Multiply by 3 before removing the logarithms.

Answer: (b) \(x=1\)

Explanation:
1. \(\dfrac{x}{\sqrt{x-1/2}}=\dfrac{x+1/2}{\sqrt{x+1/8}}\); squaring: \(x^2\left(x+\dfrac18\right)=\left(x+\dfrac12\right)^2\left(x-\dfrac12\right)\)
2. This simplifies to \(3x^2-2x-1=0\), so \(x=1\) or \(-\dfrac13\)
3. Domain needs \(x>\dfrac12\), so only \(x=1\)

Teacher's Note:
1. Rewrite the half-logs as square roots before squaring.
2. Always test each root against the domain of the original equation.

Answer: (a) \(x=100\)

Explanation:
1. \(3^{\log_3\log\sqrt x}=\log\sqrt x=\dfrac12\log x\). Put \(t=\log x\): \(\dfrac t2-t+t^2-3=0\), i.e. \(2t^2-t-6=0\)
2. \(t=2\) or \(t=-\dfrac32\)
3. \(\log_3\log\sqrt x\) needs \(\log\sqrt x>0\), so \(t>0\): \(t=2\) and \(x=100\)

Teacher's Note:
1. The identity \(a^{\log_aN}=N\) simplifies the first term.
2. Check the positivity of the inner logarithm.

Answer: (d) \(x=-2-\sqrt{10}\)

Explanation:
1. \(9^{\log_3(1-2x)}=(1-2x)^2\), so \((1-2x)^2=5x^2-5\), i.e. \(x^2+4x-6=0\)
2. \(x=-2\pm\sqrt{10}\)
3. Domain: \(1-2x>0\), so \(x<\dfrac12\); the root \(-2+\sqrt{10}\approx1.16\) is rejected

Teacher's Note:
1. Use \(9^{\log_3u}=u^2\).
2. Always test each root against the domain of the original equation.

Answer: (a) \(x=10\) and \(x=10^{-4}\)

Explanation:
1. Take \(\log\) with \(t=\log x\): \(\dfrac{t(t+7)}4=t+1\)
2. \(t^2+3t-4=0\), so \(t=1\) or \(t=-4\)
3. \(x=10\) or \(x=10^{-4}\)

Teacher's Note:
1. Multiply both sides by 4 before factorising.
2. Always test each root against the domain of the original equation.

Answer: (d) \(x=2\) and \(x=16\)

Explanation:
1. \(\log_28x=3+\log_2x\). Put \(t=\sqrt{\log_2x}\): \(3t-3-t^2+1=0\), so \(t^2-3t+2=0\)
2. \(t=1\) or \(2\), so \(\log_2x=1\) or \(4\)
3. \(x=2\) or \(16\)

Teacher's Note:
1. Substitute the square root of the logarithm as a new variable.
2. Check that the square root exists: log_2 x must be non-negative.