JEE Mathematics Logarithms MCQs Set 06

Answer: (a) \(3\)

Explanation:
1. \(\dfrac{\log_224}{\log_{96}2}=\log_224\cdot\log_296\) and \(\dfrac{\log_2192}{\log_{12}2}=\log_2192\cdot\log_212\)
2. Put \(t=\log_23\): \(\log_224=3+t,\ \log_296=5+t,\ \log_2192=6+t,\ \log_212=2+t\)
3. Difference \(=(3+t)(5+t)-(6+t)(2+t)=(15+8t+t^2)-(12+8t+t^2)=3\)

Teacher's Note:
1. Write each number as a power of 2 times a power of 3.
2. The terms in t cancel.

Answer: (d) \(469\)

Explanation:
1. \(a^{(\log_37)^2}=\left(a^{\log_37}\right)^{\log_37}=27^{\log_37}=7^{3}=343\)
2. \(b^{(\log_711)^2}=49^{\log_711}=11^2=121\)
3. \(c^{(\log_{11}25)^2}=(\sqrt{11})^{\log_{11}25}=25^{1/2}=5\)
4. Sum \(=343+121+5=469\)

Teacher's Note:
1. Split the squared exponent so that the given power can be substituted.
2. Then use a^{log_b c} = c^{log_b a}.

Answer: (c) \(x=5\)

Explanation:
1. \(2\log(x-3)=\log(x^2-21)\), so \((x-3)^2=x^2-21\)
2. \(-6x+9=-21\), so \(x=5\)

Teacher's Note:
1. Cross-multiply and remove the logarithms.
2. Always test each root against the domain of the original equation.

Answer: (a) \(x=2^{\sqrt2}\) and \(x=2^{-\sqrt2}\)

Explanation:
1. With \(t=\log_2x\): \(\dfrac1t\cdot\dfrac1{1+t}=\dfrac1{2+t}\)
2. \(2+t=t(1+t)\), so \(t^2=2\) and \(t=\pm\sqrt2\)
3. \(x=2^{\pm\sqrt2}\)

Teacher's Note:
1. Use log_x 2 = 1/log_2 x for each term.
2. Neither base equals 1 for these roots.

Answer: (d) \(507\)

Explanation:
1. Let \(t=\log_yx\): \(t+\dfrac1t=\dfrac{10}3\), so \(t=3\) or \(\dfrac13\); hence \(x=y^3\) or \(y=x^3\)
2. Take \(x=y^3\): \(y^4=144\), so \(y=2\sqrt3\) and \(x=24\sqrt3\) (the case \(y=x^3\) only swaps the roles)
3. \(x+y=26\sqrt3\), so \(\dfrac{x+y}2=13\sqrt3=\sqrt{507}\)

Teacher's Note:
1. The other case t = 1/3 just swaps x and y.
2. Square 13 root 3 to find N.

Answer: (c) \(\bar3.5386\)

Explanation:
1. \(0.003456=34.56\times10^{-4}\)
2. \(\log=1.5386-4=-2.4614\)
3. In characteristic-mantissa form \(-2.4614=-3+0.5386=\bar3.5386\)

Teacher's Note:
1. Moving the decimal point changes only the characteristic, not the mantissa.
2. The mantissa must stay positive: -2.4614 is written as bar 3 . 5386.

Answer: (b) \(0.3522\)

Explanation:
1. \(2.25=\dfrac94=\dfrac{3^2}{2^2}\)
2. \(\log2.25=2(0.4771)-2(0.3010)=0.9542-0.6020=0.3522\)

Teacher's Note:
1. Write the decimal as a ratio of prime powers.
2. Subtract, do not add, for the denominator.

Answer: (d) \(140\)

Explanation:
1. \(\log5^{200}=200(1-0.3010)=139.8\)
2. Characteristic \(=139\), so digits \(=140\)

Teacher's Note:
1. Write log 5 as 1 - log 2.
2. Digits = characteristic + 1.

Answer: (c) \(12\)

Explanation:
1. \(\log6^{15}=15(0.3010+0.4771)=11.6715\)
2. Characteristic \(=11\), so digits \(=12\)

Teacher's Note:
1. Add log 2 and log 3 to get log 6.
2. Digits = characteristic + 1.

Answer: (b) \(47\)

Explanation:
1. \(\log3^{-100}=-47.71=-48+0.29\)
2. Characteristic \(=-48\), so there are \(48-1=47\) zeros

Teacher's Note:
1. The characteristic -k means k - 1 zeros after the decimal point.
2. Write the logarithm with a positive mantissa first.

Answer: (a) \(23040\)

Explanation:
1. \(L=1024^{0.4}=(2^{10})^{0.4}=2^4=16\)
2. \(\log6^{10}=7.781\), so \(M=8\)
3. \(N=6^3-6^2=216-36=180\)
4. \(LMN=16\times8\times180=23040\)

Teacher's Note:
1. Evaluate each of the three quantities separately.
2. Characteristic 2 in base 6 means 36 up to 215.

Answer: (d) \(x=2\) and \(x=\dfrac1{32}\)

Explanation:
1. Take \(\log_2\): \((t+4)t=5\) with \(t=\log_2x\), so \(t^2+4t-5=0\)
2. \(t=1\) or \(-5\)
3. \(x=2\) or \(2^{-5}=\dfrac1{32}\)

Teacher's Note:
1. Take logarithms of both sides to bring the exponent down.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. \(x-x\log5=x\log2\), so \(x\log2+\log(1+2^x)=\log6\)
2. \(\log\left[2^x(1+2^x)\right]=\log6\), so \(y(1+y)=6\) with \(y=2^x\)
3. \(y=2\) (the root \(-3\) is impossible), so \(x=1\)

Teacher's Note:
1. Use log 10 = 1 so that x = x log 5 + x log 2.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. Put \(L=\log x\): \(5^L\left(1+\dfrac15\right)=3^L\left(3+\dfrac13\right)\)
2. \(\dfrac65\cdot5^L=\dfrac{10}3\cdot3^L\), so \(\left(\dfrac53\right)^L=\dfrac{25}9\)
3. \(L=2\), so \(x=100\)

Teacher's Note:
1. Collect powers of 5 on one side and powers of 3 on the other.
2. Compare with (5/3)^2.

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

Explanation:
1. \(1+\log_2(x-4)=\log_22(x-4)\), and \(\log_{\sqrt2}\left(\sqrt{x+3}-\sqrt{x-3}\right)=\log_2\left(\sqrt{x+3}-\sqrt{x-3}\right)^2\)
2. So \(2(x-4)=\left(\sqrt{x+3}-\sqrt{x-3}\right)^2=2x-2\sqrt{x^2-9}\), giving \(\sqrt{x^2-9}=4\)
3. \(x^2=25\), and the domain \(x>4\) gives \(x=5\)

Teacher's Note:
1. The domain is x > 4 because of log_2(x - 4).
2. Always test each root against the domain of the original equation.

Answer: (b) \(2^9\)

Explanation:
1. With \(a=\log_2x,\ b=\log_2y\): \(\dfrac a3+b=5\) and \(\dfrac b3+a=7\)
2. Adding: \(\dfrac43(a+b)=12\), so \(a+b=9\)
3. \(\log_2xy=9\), so \(xy=2^9\)

Teacher's Note:
1. Add the two equations rather than solving for each variable.
2. Convert all logarithms to base 2.

Answer: (a) \(1\)

Explanation:
1. Let \(p=\log_23,\ q=\log_{3}2\). \(a=\dfrac{1+2p}{2+p}\) and \(b=\dfrac{1+3p}{3+p}\)
2. \(ab+5(a-b)=\dfrac{(1+2p)(1+3p)+5(1+2p)(3+p)-5(1+3p)(2+p)}{(2+p)(3+p)}\)
3. Numerator \(=6p^2+5p+1+5(2p^2+7p+3)-5(3p^2+7p+2)=p^2+5p+6=(2+p)(3+p)\), so the value is 1

Teacher's Note:
1. Write both logarithms with log_2 3 = p.
2. The numerator factorises exactly into the denominator.

Answer: (d) \(x=2\) and \(x=1-\sqrt{33}\)

Explanation:
1. \(\log_4|x+2|^3+3=\log_4(4-x)^3(6+x)^3\), so \(|x+2|\cdot4=(4-x)(6+x)\) with domain \(-6<x<4\)
2. If \(x>-2\): \(4x+8=-x^2-2x+24\), so \(x^2+6x-16=0\), \(x=2\) (or \(-8\), rejected)
3. If \(x<-2\): \(-4x-8=-x^2-2x+24\), so \(x^2-2x-32=0\), \(x=1-\sqrt{33}\approx-4.74\)

Teacher's Note:
1. Use |x + 2| because (x + 2)^2 is always positive.
2. Split into the cases x > -2 and x < -2.

Answer: (c) \((2008)^2\)

Explanation:
1. Take \(\log_{2008}\) and put \(t=\log_{2008}x\): \(\dfrac12+t^2=2t\), so \(2t^2-4t+1=0\)
2. Sum of the values of t \(=2\)
3. Product of roots \(=2008^{t_1}\cdot2008^{t_2}=2008^{t_1+t_2}=2008^2\)

Teacher's Note:
1. The product of the roots x1 x2 equals the base raised to the sum of the t-values.
2. Sum of roots of a quadratic is -b/a.

Answer: (b) \((-10,20)\) and \(\left(\dfrac{10}3,\dfrac{20}3\right)\)

Explanation:
1. \(|x+y|=100^{1/2}=10\) and \(\log_{10}\dfrac y{|x|}=\log_{10}2\), so \(y=2|x|\) (with \(y>0\))
2. If \(x>0\): \(x+2x=10\), so \(x=\dfrac{10}3,\ y=\dfrac{20}3\)
3. If \(x<0\): \(x+2(-x)=-x=\pm10\); since \(-x>0\), \(x=-10\) and \(y=20\)

Teacher's Note:
1. Both signs of x must be examined because of the modulus.
2. y is positive since log y is defined, so x + y = -10 is impossible.

Answer: (a) \(x=8\)

Explanation:
1. \(\log_{1/2}u=-\log_2u\), so the equation is \(\log_2\dfrac{(\log_2x)^2}{\log_2(2\sqrt2x)}=1\)
2. With \(t=\log_2x\): \(t^2=2\left(t+\dfrac32\right)\), so \(t^2-2t-3=0\), \(t=3\) or \(-1\)
3. \(\log_2(\log_2x)\) needs \(t>0\), so \(t=3\) and \(x=8\)

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

Answer: (c) \(\dfrac12\)

Explanation:
1. Second equation: \(\ln x\ln3=\ln y\ln2\), so \(\ln y=\dfrac{\ln3}{\ln2}\ln x\)
2. First equation: \(\ln2(\ln2+X)=\ln3(\ln3+\ln y)\) with \(X=\ln x\), i.e. \(\ln^22+X\ln2=\ln^23+\dfrac{\ln^23}{\ln2}X\)
3. \(X\,\dfrac{\ln^22-\ln^23}{\ln2}=\ln^23-\ln^22\), so \(X=-\ln2\) and \(x_0=\dfrac12\)

Teacher's Note:
1. Take natural logarithms of both equations to get linear equations in ln x and ln y.
2. Check: y = 2^(-ln3/ln2 ...) gives (2x)^{ln 2} = 1 and (3y)^{ln 3} = 1.

Answer: (b) \(8\)

Explanation:
1. Let \(m=\log_29\). \(\dfrac1{\log_2m}=\log_m2\), so \((m^2)^{\log_m2}=\left(m^{\log_m2}\right)^2=2^2=4\)
2. \(\dfrac1{\log_47}=\log_74\), so \((\sqrt7)^{\log_74}=\left(7^{\log_74}\right)^{1/2}=4^{1/2}=2\)
3. Product \(=4\times2=8\)

Teacher's Note:
1. Reciprocal of a logarithm flips base and argument.
2. Then use a^{log_a N} = N.