JEE Mathematics Logarithms MCQs Set 04

Answer: (b) \(x=\dfrac13\) and \(x=\dfrac1{81}\)

Explanation:
1. Put \(t=\sqrt{\log_{1/3}x}\): \(t^2-3t+2=0\), so \(t=1\) or \(2\)
2. \(\log_{1/3}x=1\) or \(4\)
3. \(x=\dfrac13\) or \(\left(\dfrac13\right)^4=\dfrac1{81}\)

Teacher's Note:
1. The square root needs log_{1/3} x to be non-negative, i.e. x at most 1.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. \(2-\log_48=\log_416-\log_48=\log_42\)
2. \(\dfrac{x+3}{x-1}=2\), so \(x+3=2x-2\) and \(x=5\)

Teacher's Note:
1. Write 2 as log_4 16 so that all terms are logarithms.
2. Always test each root against the domain of the original equation.

Answer: (b) \(x=-4\)

Explanation:
1. \(2\log_4(4-x)=\log_2(4-x)\), so \(\log_2(4-x)+\log_2(-2-x)=4\)
2. \((4-x)(-2-x)=16\), i.e. \(x^2-2x-24=0\), so \(x=6\) or \(-4\)
3. Domain needs \(-2-x>0\), so \(x<-2\): only \(x=-4\)

Teacher's Note:
1. Convert log base 4 into base 2 first.
2. Always test each root against the domain of the original equation.

Answer: (a) \(x=8\) and \(x=\dfrac{\sqrt3}{3}-1\)

Explanation:
1. Put \(t=\log_3(x+1)\); then \(\log_{x+1}3=\dfrac1t\): \(3+\dfrac2t=2t\), so \(2t^2-3t-2=0\)
2. \(t=2\) or \(t=-\dfrac12\)
3. \(x+1=9\) or \(3^{-1/2}\), so \(x=8\) or \(x=\dfrac1{\sqrt3}-1=\dfrac{\sqrt3}3-1\)

Teacher's Note:
1. The reciprocal rule changes log_{x+1} 3 to 1/t.
2. Both roots keep the base x + 1 positive and different from 1.

Answer: (d) \(x=0\) and \(x=-1\)

Explanation:
1. \(4\cdot3^x-1=3^{2x+1}=3y^2\) with \(y=3^x\), so \(3y^2-4y+1=0\)
2. \(y=1\) or \(y=\dfrac13\), i.e. \(x=0\) or \(x=-1\)
3. Both give a positive argument (3 and \(\dfrac13\))

Teacher's Note:
1. Substitute y = 3^x and solve the quadratic.
2. Always test each root against the domain of the original equation.

Answer: (a) \(x=\dfrac1{16}\) and \(x=2\)

Explanation:
1. The argument must equal \(3^2=9\). Put \(t=\log_{1/2}x\): \(t^2-3t+5=9\), so \(t^2-3t-4=0\)
2. \(t=4\) or \(t=-1\)
3. \(x=\left(\dfrac12\right)^4=\dfrac1{16}\) or \(x=\left(\dfrac12\right)^{-1}=2\)

Teacher's Note:
1. Remove the outer logarithm first, then substitute.
2. The expression under the log is always positive here (discriminant negative), so no domain loss.

Answer: (c) \(x=23\) and \(x=-\dfrac95\)

Explanation:
1. Put \(t=\log_5(x+2)\); \(\log_{x+2}5=\dfrac1t\): \(1+\dfrac2t=t\), so \(t^2-t-2=0\)
2. \(t=2\) or \(-1\), i.e. \(x+2=25\) or \(\dfrac15\)
3. \(x=23\) or \(x=-\dfrac95\)

Teacher's Note:
1. Both values of x+2 are positive and different from 1.
2. Check the base condition when the base contains x.

Answer: (b) \(x=2\)

Explanation:
1. \(\log_42^{4x}=4x\log_42=2x\)
2. \(2^{\log_24}=4\)
3. So \(2x=4\) and \(x=2\)

Teacher's Note:
1. Use the power rule on the left and a^{log_a N} = N on the right.
2. Always simplify each side separately.

Answer: (b) \(x=2\)

Explanation:
1. \(2\cdot4^{x-2}-1=4^{2x-4}\). Put \(y=4^{x-2}\): \(2y-1=y^2\)
2. \((y-1)^2=0\), so \(y=1\) and \(x=2\)

Teacher's Note:
1. Isolate the logarithm, then use the definition.
2. A perfect square gives only one root.

Answer: (a) \(x=5\) and \(x=\sqrt[5]5\)

Explanation:
1. Put \(u=\log_x\sqrt5\); \(\log_x5x=1+2u\): \(u+1+2u-2.25=u^2\), so \(u^2-3u+1.25=0\)
2. \(u=\dfrac12\) or \(\dfrac52\), i.e. \(\log_x5=1\) or \(5\)
3. \(x=5\) or \(x=5^{1/5}\)

Teacher's Note:
1. Express log_x(5x) as 1 + log_x 5 = 1 + 2u.
2. Always test each root against the domain of the original equation.

Answer: (d) \(x=10\)

Explanation:
1. \(t=\log x\): \(\log[t(4t-3)]=0\), so \(t(4t-3)=1\)
2. \(4t^2-3t-1=0\), so \(t=1\) or \(-\dfrac14\)
3. \(\log(\log x)\) needs \(t>0\), so \(t=1\) and \(x=10\)

Teacher's Note:
1. Nested logarithms require the inner logarithm to be positive.
2. Always test each root against the domain of the original equation.

Answer: (b) \(x=4\)

Explanation:
1. \(\log x^2=\log(5x-4)\), so \(x^2-5x+4=0\) and \(x=1\) or \(4\)
2. At \(x=1\), \(\log(5x-4)=\log1=0\), so the denominator vanishes
3. Hence \(x=4\)

Teacher's Note:
1. Check the denominator for each root.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. \(\log_{1/8}\sqrt{3x-5}=-\dfrac16\log_2(3x-5)\), so \(\log_2(x-2)-2=-\log_2(3x-5)\)
2. \(\log_2[(x-2)(3x-5)]=2\), so \(3x^2-11x+6=0\), \(x=3\) or \(\dfrac23\)
3. Domain needs \(x>2\), so \(x=3\)

Teacher's Note:
1. Multiply the whole equation by 6 to clear the fractions.
2. Always test each root against the domain of the original equation.

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

Explanation:
1. Put \(t=\log_2x\): \(\log_22x^2=1+2t\), \(\log_416x=\dfrac{4+t}2\), \(\log_4x^3=\dfrac{3t}2\)
2. Squaring: \(\dfrac{(1+2t)(4+t)}2=\dfrac{9t^2}4\), so \(5t^2-18t-8=0\), \(t=4\) or \(-\dfrac25\)
3. The right side must be non-negative, so \(t\ge0\): \(t=4\) and \(x=16\)

Teacher's Note:
1. A square root equals a non-negative number, which eliminates a root after squaring.
2. Always test each root against the domain of the original equation.

Answer: (c) \(x=100\) and \(x=10^{-5/3}\)

Explanation:
1. Put \(t=\log x\): \(3t+19=(2t+1)(3t-1)=6t^2+t-1\)
2. \(3t^2-t-10=0\), so \(t=2\) or \(t=-\dfrac53\)
3. \(x=100\) or \(x=10^{-5/3}\); neither makes \(3t-1\) zero

Teacher's Note:
1. Cross-multiply after substituting t.
2. Always test each root against the domain of the original equation.

Answer: (b) \(x=48\)

Explanation:
1. \(3\log\sqrt[3]{x-40}=\log(x-40)\), so \(\sqrt{x+1}+1=x-40\)
2. \(\sqrt{x+1}=x-41\); squaring, \(x^2-83x+1680=0\), so \(x=48\) or \(35\)
3. \(x-41\ge0\) is needed, so \(x=35\) is extraneous and \(x=48\)

Teacher's Note:
1. Squaring can introduce an extraneous root, so test both values.
2. Check that the right side x - 41 is positive.

Answer: (a) \(x=10^{\sqrt3}\) and \(x=10^{-\sqrt3}\)

Explanation:
1. \(\log_3^26-\log_3^22=(\log_36-\log_32)(\log_36+\log_32)=1\cdot\log_312\)
2. So \(\log_312=(\log^2x-2)\log_312\), giving \(\log^2x-2=1\)
3. \(\log x=\pm\sqrt3\), so \(x=10^{\pm\sqrt3}\)

Teacher's Note:
1. Use the difference of squares to evaluate the left side.
2. Both roots are positive, so both are valid.

Answer: (d) \(x=\dfrac1{\sqrt2}\), \(x=1\) and \(x=4\)

Explanation:
1. Put \(t=\log_2x\). Then \(\dfrac{2t}{t-1}-\dfrac{42t}{t+4}+\dfrac{20t}{t+2}=0\)
2. Either \(t=0\), or \(\dfrac2{t-1}-\dfrac{42}{t+4}+\dfrac{20}{t+2}=0\), which simplifies to \(2t^2-3t-2=0\), so \(t=2\) or \(t=-\dfrac12\)
3. So \(x=2^0=1\), \(x=2^2=4\), \(x=2^{-1/2}=\dfrac1{\sqrt2}\)

Teacher's Note:
1. Convert every logarithm to base 2.
2. The bases 0.5x, 16x and 4x must not equal 1, which holds for all three roots.

Answer: (c) \(\log_8d<\log_3d<\log_ed<\log_2d\)

Explanation:
1. For \(d>1\), \(\log_bd=\dfrac{\ln d}{\ln b}\) with \(\ln d>0\)
2. A larger base gives a larger denominator and hence a smaller logarithm
3. Bases in increasing order: \(2<e<3<8\), so the logarithms decrease: \(\log_2d>\log_ed>\log_3d>\log_8d\)

Teacher's Note:
1. For numbers above 1, a bigger base gives a smaller logarithm.
2. Remember that e is about 2.718, which lies between 2 and 3.

Answer: (b) \(0.954\)

Explanation:
1. \(\log27=3\log3=1.431\), so \(\log3=0.477\)
2. \(\log9=2\log3=0.954\)

Teacher's Note:
1. Find log 3 first; both 27 and 9 are powers of 3.
2. Use the power rule for logarithms.

Answer: (a) \(2yz\)

Explanation:
1. \(xyz=\log_{2a}a\cdot\log_{3a}2a\cdot\log_{4a}3a=\log_{4a}a\) (changing everything to base \(4a\))
2. \(xyz+1=\log_{4a}a+\log_{4a}4a=\log_{4a}4a^2=\log_{4a}(2a)^2=2\log_{4a}2a\)
3. \(\log_{4a}2a=\log_{3a}2a\cdot\log_{4a}3a=yz\), so \(xyz+1=2yz\)

Teacher's Note:
1. The chain of logarithms telescopes because each argument is the next base.
2. Write 1 as log of the base to combine terms.

Answer: (c) \(25\)

Explanation:
1. \(\log_5a\cdot\log_ax=\log_5x\) (change of base)
2. \(\log_5x=2\), so \(x=5^2=25\)

Teacher's Note:
1. The product of two chained logarithms collapses to one.
2. Here a disappears completely.

Answer: (b) \(\log2\)

Explanation:
1. Write each fraction in primes: \(\dfrac{81}{80}=\dfrac{3^4}{2^4\cdot5}\), \(\dfrac{25}{24}=\dfrac{5^2}{2^3\cdot3}\), \(\dfrac{16}{15}=\dfrac{2^4}{3\cdot5}\)
2. Power of 2: \(-12-15+28=1\); power of 3: \(12-5-7=0\); power of 5: \(-3+10-7=0\)
3. So the value is \(\log2\)

Teacher's Note:
1. Collect the exponent of each prime separately.
2. Constants like 1 or 0 for the exponents make the answer a single logarithm.

Answer: (a) \((-\infty,-1]\cup[1,\infty)\)

Explanation:
1. We need \(x^2>0\) and \(\log_{10}x^2\ge0\)
2. \(\log_{10}x^2\ge0\Rightarrow x^2\ge1\)
3. So \(x\le-1\) or \(x\ge1\)

Teacher's Note:
1. Do not forget the negative values of x because only x squared appears.
2. Square roots need a non-negative quantity inside.