JEE Mathematics Logarithms MCQs Set 05

Answer: (d) \(z\)

Explanation:
1. \(\log_xa\cdot\log_ay\cdot\log_yz=\log_xz\) (the chain of logarithms)
2. \(x^{\log_xz}=z\)

Teacher's Note:
1. Use the chain rule log_b a . log_c b = log_c a repeatedly.
2. Then apply a^{log_a N} = N.

Answer: (c) \(16\)

Explanation:
1. \(\log_2xy\ge6\Rightarrow xy\ge64\) (with x, y positive)
2. By AM-GM, \(x+y\ge2\sqrt{xy}\ge2\times8=16\)
3. Equality holds at \(x=y=8\)

Teacher's Note:
1. Combine the logarithms and use AM-GM on the product.
2. Check that equality is reached.

Answer: (b) \(100\)

Explanation:
1. We need \(N=50\log_{10}N\)
2. Try \(N=100\): \(50\log100=50\times2=100\), which is true
3. Since \(\log_{10}N=2\) is rational only for powers of 10, \(N=100\)

Teacher's Note:
1. Test powers of 10 because only they have rational logarithms among rationals.
2. Verify by substitution.

Answer: (a) \(x>y\)

Explanation:
1. \(5^4=625<1000\), so \(x>4\)
2. \(7^4=2401>2058\), so \(y<4\)
3. Hence \(x>4>y\)

Teacher's Note:
1. Compare both with the integer 4.
2. Compare powers rather than computing decimals.

Answer: (d) \(10\)

Explanation:
1. \(4^{\log_93}=4^{1/2}=2\) and \(9^{\log_24}=9^2=81\)
2. So \(10^{\log_x83}=83\), i.e. \(\log_x83=\log_{10}83\)
3. Hence \(x=10\)

Teacher's Note:
1. Evaluate the left side to a plain number.
2. Equal logarithms of the same number have equal bases.

Answer: (c) \(7\)

Explanation:
1. \(\dfrac{a+b}3=\sqrt{ab}\), so \((a+b)^2=9ab\)
2. \(a^2+b^2=7ab\)
3. \(\dfrac ab+\dfrac ba=\dfrac{a^2+b^2}{ab}=7\)

Teacher's Note:
1. Remove the logarithms by exponentiating, then square.
2. Divide by ab at the end.

Answer: (d) \((-\infty,0)\cup\left(\dfrac85,\infty\right)\)

Explanation:
1. \(\log_2|4-5x|>2\Rightarrow|4-5x|>4\)
2. \(4-5x>4\Rightarrow x<0\), or \(4-5x<-4\Rightarrow x>\dfrac85\)
3. Solution: \(x<0\) or \(x>\dfrac85\)

Teacher's Note:
1. Base 2 is greater than 1, so the inequality direction is preserved.
2. An absolute value greater than 4 splits into two cases.

Answer: (c) \(\left[\dfrac1{100},\dfrac1{\sqrt{10}}\right]\)

Explanation:
1. \(\log_{0.1}x=-\log_{10}x\), so \(\dfrac12\le-\log x\le2\)
2. \(-2\le\log x\le-\dfrac12\)
3. \(\dfrac1{100}\le x\le10^{-1/2}=\dfrac1{\sqrt{10}}\)

Teacher's Note:
1. Base 0.1 is less than 1, so the inequality reverses when you remove the logarithm.
2. Convert to base 10 to avoid mistakes.

Answer: (b) \(x=\dfrac{484}{49}\)

Explanation:
1. \(\log_7(\sqrt{x+5}+\sqrt x)=11^0=1\), so \(\sqrt{x+5}+\sqrt x=7\)
2. \(\sqrt{x+5}=7-\sqrt x\Rightarrow x+5=49-14\sqrt x+x\), so \(\sqrt x=\dfrac{22}7\)
3. \(x=\dfrac{484}{49}\)

Teacher's Note:
1. Peel off the outer logarithm first and then the inner one.
2. Isolate one square root before squaring.

Answer: (d) \(12\)

Explanation:
1. The two numbers inside the logarithms are the roots of \(t^2-abt+(a+b)=0\); their product is \(a+b=100\)
2. \(A=\log_{10}100=2\)
3. Let \(p=\log_62,\ q=\log_63\), so \(p+q=1\) and \(\log_618=p+2q=1+q\). Then \(\log_6B=(1+q)p+q\cdot q=p+q(p+q)=p+q=1\), so \(B=6\)
4. \(A\cdot B=2\times6=12\)

Teacher's Note:
1. The sum and product of conjugate expressions often reduce to a simple number.
2. Evaluate B by writing 18 = 2 x 9.

Answer: (c) \(\dfrac{xy+2}{2y}\)

Explanation:
1. \(\log_310=\log_32+\log_35\)
2. \(\log_32=\dfrac x2\) and \(\log_35=\dfrac1y\)
3. \(\log_310=\dfrac x2+\dfrac1y=\dfrac{xy+2}{2y}\)

Teacher's Note:
1. Express every logarithm through x and y by taking reciprocals where needed.
2. Split 10 as 2 x 5.

Answer: (b) \(\dfrac{xy+2y-2}{2y}\)

Explanation:
1. \(1.2=\dfrac65\), so \(\log_31.2=\log_32+\log_33-\log_35\)
2. \(=\dfrac x2+1-\dfrac1y=\dfrac{xy+2y-2}{2y}\)

Teacher's Note:
1. Write 1.2 as 6/5 and split 6 into 2 x 3.
2. Keep the sign of log_3 5 negative.

Answer: (a) \(625\)

Explanation:
1. \(k^{(\log_25)^2}=\left(k^{\log_25}\right)^{\log_25}=16^{\log_25}\)
2. \(16^{\log_25}=2^{4\log_25}=5^4=625\)

Teacher's Note:
1. Write the exponent as a product so the given value can be substituted.
2. Use a^{log_b c} = c^{log_b a}.

Answer: (d) \(x=16\) and \(x=-4\)

Explanation:
1. \(x^2-12x+36=(x-6)^2=100\)
2. \(x-6=\pm10\), so \(x=16\) or \(-4\)
3. Both give \((x-6)^2=100>0\)

Teacher's Note:
1. Recognise the perfect square before expanding.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. \(3^{1+\log_3x}=3x\), so \(9x^2-3x-210=0\), i.e. \(3x^2-x-70=0\)
2. \(x=5\) or \(-\dfrac{14}3\)
3. \(\log_3x\) needs \(x>0\), so \(x=5\)

Teacher's Note:
1. Use a^{log_a N} = N to simplify 3^{1+log_3 x}.
2. Always test each root against the domain of the original equation.

Answer: (d) \(3\)

Explanation:
1. \(\log_5[\sqrt{2x-2}+3]=e^0=1\)
2. \(\sqrt{2x-2}+3=5\), so \(\sqrt{2x-2}=2\)
3. \(2x-2=4\), so \(x=3\)

Teacher's Note:
1. Remove ln first, then log base 5.
2. Check by substituting back.

Answer: (c) 2 is smaller

Explanation:
1. Let \(u=\log_2\pi>0\); then \(\log_\pi2=\dfrac1u\)
2. \(u+\dfrac1u\ge2\) by AM-GM, with equality only if \(u=1\), i.e. \(\pi=2\)
3. Since \(\pi\ne2\), \(\log_\pi2+\log_2\pi>2\), so 2 is smaller

Teacher's Note:
1. The sum of a positive number and its reciprocal is at least 2.
2. Equality would need the base and number to be equal.

Answer: (b) \(2\)

Explanation:
1. Let \(t=\log_ab\) and \(u=\log_ac\ne0\). Then \(\log_bc=\dfrac ut\) and \(\log_{ab}c=\dfrac u{1+t}\)
2. \(2u\left(1+\dfrac1t\right)=\dfrac{9u}{1+t}\), so \(2(1+t)^2=9t\), i.e. \(2t^2-5t+2=0\)
3. \(t=2\) or \(\dfrac12\); the largest value is 2

Teacher's Note:
1. Express all three logarithms through log_a b and log_a c.
2. Cancel u only because c is not 1.

Answer: (a) \(3721\)

Explanation:
1. Put \(p=\ln x\). Dividing by \(\dfrac{1}{\ln3\ln4\ln5}\): \(p^3=p^2(\ln3+\ln4+\ln5)\)
2. So \(p=0\) or \(p=\ln60\), i.e. \(x=1\) or \(x=60\)
3. Sum \(=61\), and its square is \(3721\)

Teacher's Note:
1. Do not divide by p^2 blindly: p = 0 is also a root.
2. Both x = 1 and x = 60 satisfy the original equation.

Answer: (d) \(\dfrac16\)

Explanation:
1. \(\dfrac2{6\log_42000}=\dfrac{\log_{2000}4}{3}\) and \(\dfrac3{6\log_52000}=\dfrac{\log_{2000}5}2\)
2. Sum \(=\dfrac16\left(2\log_{2000}4+3\log_{2000}5\right)=\dfrac16\log_{2000}(16\times125)\)
3. \(=\dfrac16\log_{2000}2000=\dfrac16\)

Teacher's Note:
1. Bring the power 6 down and reciprocate each logarithm.
2. Notice that 16 x 125 = 2000.

Answer: (b) \(6\)

Explanation:
1. \(5^{\log_{1/5}(1/2)}=5^{\log_52}=2\)
2. \(\log_{\sqrt2}\dfrac4{\sqrt7+\sqrt3}=2\log_24-2\log_2(\sqrt7+\sqrt3)\) and \(10+2\sqrt{21}=(\sqrt7+\sqrt3)^2\)
3. \(\log_{1/2}\dfrac1{10+2\sqrt{21}}=2\log_2(\sqrt7+\sqrt3)\). Total \(=2+4-2L+2L=6\)

Teacher's Note:
1. Recognise 10 + 2 root 21 as a perfect square.
2. The two remaining logarithms cancel except for log_2 4 terms.

Answer: (a) \(x=\dfrac1{100}\)

Explanation:
1. Put \(L=\log_{10}x\): \(4\cdot4^L-6^L-18\cdot9^L=0\). Divide by \(9^L\): \(4\left(\dfrac49\right)^L-\left(\dfrac23\right)^L-18=0\)
2. With \(w=\left(\dfrac23\right)^L\): \(4w^2-w-18=0\), so \(w=\dfrac94\) (the other root is negative)
3. \(\left(\dfrac23\right)^L=\left(\dfrac23\right)^{-2}\), so \(L=-2\) and \(x=\dfrac1{100}\)

Teacher's Note:
1. Divide by the largest base power to get a quadratic in (2/3)^L.
2. An exponential can never be negative.

Answer: (c) \(\dfrac{25}2\)

Explanation:
1. \(49^{1-\log_72}=\dfrac{49}{49^{\log_72}}=\dfrac{49}{2^2}=\dfrac{49}4\)
2. \(5^{-\log_54}=\dfrac1{5^{\log_54}}=\dfrac14\)
3. Sum \(=\dfrac{50}4=\dfrac{25}2\)

Teacher's Note:
1. Split the exponent into a quotient of powers.
2. Use a^{log_a N} = N for each term.

Answer: (b) \(\dfrac{1+2ac}{2c+abc+1}\)

Explanation:
1. \(\log_{140}63=\dfrac{\log_263}{\log_2140}\) with \(\log_263=2a+\log_27=2a+\dfrac1c\)
2. \(\log_2140=2+\log_25+\log_27=2+ab+\dfrac1c\)
3. Ratio \(=\dfrac{2a+1/c}{2+ab+1/c}=\dfrac{2ac+1}{2c+abc+1}\)

Teacher's Note:
1. Choose base 2 for everything, since all given logarithms connect to it.
2. Write 63 = 9 x 7 and 140 = 4 x 5 x 7.