Download important IIFT Logarithms Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Logarithms questions and answers for IIFT exam.<\/p>\n
Download Logarithms Questions for IIFT PDF\u00a0<\/a><\/p>\n Get 50% Off on IIFT & Other MBA Test Series\u00a0<\/a><\/p>\n Download IIFT Previous Papers PDF<\/a><\/p>\n Practice IIFT Mock Tests<\/a><\/p>\n Question 1:\u00a0<\/b>If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7\/2)$ are in arithmetic progression, then the value of x is equal to<\/p>\n a)\u00a05<\/p>\n b)\u00a04<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n Question 2:\u00a0<\/b>If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?<\/p>\n a)\u00a0(-2, 1\/2)<\/p>\n b)\u00a0(1,1)<\/p>\n c)\u00a0(0.4, 2.5)<\/p>\n d)\u00a0($\\pi$, 1\/ $\\pi$)<\/p>\n e)\u00a0(2,2)<\/p>\n Question 3:\u00a0<\/b>If $f(x) = \\log \\frac{(1+x)}{(1-x)}$, then f(x) + f(y) is<\/p>\n a)\u00a0$f(x+y)$<\/p>\n b)\u00a0$f{\\frac{(x+y)}{(1+xy)}}$<\/p>\n c)\u00a0$(x+y)f{\\frac{1}{(1+xy)}}$<\/p>\n d)\u00a0$\\frac{f(x)+f(y)}{(1+xy)}$<\/p>\n Question 4:\u00a0<\/b>If $\\log_{2}{x}.\\log_{\\frac{x}{64}}{2}=\\log_{\\frac{x}{16}}{2}$. Then x is<\/p>\n a)\u00a02<\/p>\n b)\u00a04<\/p>\n c)\u00a016<\/p>\n d)\u00a012<\/p>\n Question 5:\u00a0<\/b>Find the value of x from the following equation: a)\u00a02\/7<\/p>\n b)\u00a07\/2<\/p>\n c)\u00a09\/2<\/p>\n d)\u00a0None of the above<\/p>\n IIFT Free Mock Test<\/a><\/p>\n Enroll for CAT\/MBA Courses<\/a><\/p>\n Question 6:\u00a0<\/b>If $log_{10} x – log_{10} \\sqrt[3]{x} = 6log_{x}10$ then the value of x is<\/p>\n a)\u00a010<\/p>\n b)\u00a030<\/p>\n c)\u00a0100<\/p>\n d)\u00a01000<\/p>\n Question 7:\u00a0<\/b>$(1+5)\\log_{e}3+\\frac{(1+5^{2})}{2!}(\\log_{e}3)^{2}+\\frac{(1+5^{3})}{3!}(\\log_{e}3)^{3}+…$<\/p>\n a)\u00a012<\/p>\n b)\u00a0244<\/p>\n c)\u00a0243<\/p>\n d)\u00a0245<\/p>\n Question 8:\u00a0<\/b>If $log(2^{a}\\times3^{b}\\times5^{c} )$is the arithmetic mean of $log ( 2^{2}\\times3^{3}\\times5)$, $log(2^{6}\\times3\\times5^{7} )$, and $log(2 \\times3^{2}\\times5^{4} )$, then a equals<\/p>\n Question 9:\u00a0<\/b>If $\\log_{2}({5+\\log_{3}{a}})=3$ and $\\log_{5}({4a+12+\\log_{2}{b}})=3$, then a + b is equal to<\/p>\n a)\u00a059<\/p>\n b)\u00a040<\/p>\n c)\u00a032<\/p>\n d)\u00a067<\/p>\n Question 10:\u00a0<\/b>$\\frac{1}{log_{2}100}-\\frac{1}{log_{4}100}+\\frac{1}{log_{5}100}-\\frac{1}{log_{10}100}+\\frac{1}{log_{20}100}-\\frac{1}{log_{25}100}+\\frac{1}{log_{50}100}$=?<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a010<\/p>\n c)\u00a00<\/p>\n d)\u00a0\u22124<\/p>\n Top 500+ Free Questions for IIFT<\/a><\/p>\n IIFT Previous year question\u00a0 answer PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $2 log (2^x – 5) = log 2 + log (2^x – 7\/2)$ 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $log_y x = ab$ 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n If $f(x) = \\log \\frac{(1+x)}{(1-x)}$ then $f(y) = \\log \\frac{(1+y)}{(1-y)}$<\/p>\n Also Log (A*B)= Log A + Log B<\/p>\n f(x)+f(y) = $ \\log \\frac{(1+x)(1+y)}{(1-x)(1-y)}$ solving we get $\\log { \\frac{1+ \\frac{(x+y)}{(1+xy)}}{1- \\frac{(x+y)}{(1+xy)}}}$<\/p>\n Hence option B.<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\log_{2}{x}.\\log_{\\frac{x}{64}}{2}=\\log_{\\frac{x}{16}}{2}$<\/p>\n i.e. $\\frac{log{x}}{log{2}} * \\frac{log_{2}}{log{x}-log{64}} = \\frac{log{2}}{log{x}-log{16}}$<\/p>\n i.e.\u00a0$\\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\\log{2}$<\/p>\n let t = log x<\/p>\n Therefore,\u00a0\u00a0$\\frac{t * (t-log{16})}{t-log{64}}$ = $\\log{2}$<\/p>\n $t^2-4*log 2*t = t*log 2-6*(log 2)^2$<\/p>\n I.e.\u00a0$t^2-5*log 2*t-6*(log 2)^2$ = 0<\/p>\n I.e.\u00a0$t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0<\/p>\n i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0<\/p>\n i.e $t=2*log 2$ or $t=3*log 2$<\/p>\n i.e $log x=log 4$ or $log x=log 8$<\/p>\n therefore $x=4$ or $8$<\/p>\n therefore our answer is option ‘B’<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+1$ can be written as<\/p>\n $\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+\\log_{10}{10}$<\/p>\n We know that\u00a0$\\log_{10}{a}+\\log_{10}{b}=\\log_{10}{ab}$<\/p>\n $\\log_{10}{3*(4x+1)}=\\log_{10}{(x+1)*10}$<\/p>\n $12x+3=10x+10$<\/p>\n $x=7\/2$. Hence, option B is the correct answer.<\/p>\n Quant Formula For IIFT PDF<\/a><\/p>\n IIFT Free Mock Test<\/a><\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\log_{10} x – \\log_{10} \\sqrt[3]{x} = 6\\log_{x}10$ 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Splitting the above mentioned series into two series<\/p>\n A = $\\log_{e}3+\\frac{1}{2!}(\\log_{e}3)^{2}+\\frac{1}{3!}(\\log_{e}3)^{3}+…$<\/p>\n B = $5\\log_{e}3+\\frac{5^{2}}{2!}(\\log_{e}3)^{2}+\\frac{5^{3}}{3!}(\\log_{e}3)^{3}+…$<\/p>\n We know that $e^{x}$ =$1+x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+…$<\/p>\n So\u00a0\u00a0$e^{x}-1$ = $x+\\frac{x^{2}}{2!}+\\frac{x^{3}}{3!}+…$<\/p>\n On solving two series A and B<\/p>\n A = $\\log_{e}3+\\frac{1}{2!}(\\log_{e}3)^{2}+\\frac{1}{3!}(\\log_{e}3)^{3}+…$ =$e^{\\log_{e}3}-1$ = $3-1$ =$2$<\/p>\n B = $5\\log_{e}3+\\frac{5^{2}}{2!}(\\log_{e}3)^{2}+\\frac{5^{3}}{3!}(\\log_{e}3)^{3}+…$<\/span>=$e^{\\log_{e}3^{5}}-1$=$3^{5}-1$=$242$<\/p>\n A+B = $2 + 242$ = $244$<\/p>\n 8)\u00a0Answer:\u00a03<\/b><\/p>\n $log(2^{a}\\times3^{b}\\times5^{c} )$ = $ \\frac{log ( 2^{2}\\times3^{3}\\times5) + log(2^{6}\\times3\\times5^{7} ) + log(2 \\times3^{2}\\times5^{4} ) }{3} $<\/p>\n $log(2^{a}\\times3^{b}\\times5^{c} )$ = $ \\frac{log ( 2^{2+6+1}\\times3^{3+1+2}\\times5^{1+7+4}) }{3} $<\/p>\n $log(2^{a}\\times3^{b}\\times5^{c} )$ = $ \\frac{log ( 2^{9}\\times3^{6}\\times5^{12}) }{3} $<\/p>\n $3log(2^{a}\\times3^{b}\\times5^{c} )$ = $ log ( 2^{9}\\times3^{6}\\times5^{12}) $ 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\log_{2}({5+\\log_{3}{a}})=3$ $\\log_{5}({4a+12+\\log_{2}{b}})=3$ So, $a + b$ = 27 + 32 = 59 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We know that\u00a0$\\dfrac{1}{log_{a}{b}}$ = $\\dfrac{log_{x}{a}}{log_{x}{b}}$<\/p>\n Therefore, we can say that\u00a0$\\dfrac{1}{log_{2}{100}}$ = $\\dfrac{log_{10}{2}}{log_{10}{100}}$<\/p>\n $\\Rightarrow$ $\\frac{1}{log_{2}100}-\\frac{1}{log_{4}100}+\\frac{1}{log_{5}100}-\\frac{1}{log_{10}100}+\\frac{1}{log_{20}100}-\\frac{1}{log_{25}100}+\\frac{1}{log_{50}100}$<\/p>\n $\\Rightarrow$ $\\dfrac{log_{10}{2}}{log_{10}{100}}$-$\\dfrac{log_{10}{4}}{log_{10}{100}}$+$\\dfrac{log_{10}{5}}{log_{10}{100}}$-$\\dfrac{log_{10}{10}}{log_{10}{100}}$+$\\dfrac{log_{10}{20}}{log_{10}{100}}$-$\\dfrac{log_{10}{25}}{log_{10}{100}}$+$\\dfrac{log_{10}{50}}{log_{10}{100}}$<\/p>\n We know that $log_{10}{100}=2$<\/p>\n $\\Rightarrow$ $\\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$<\/p>\n $\\Rightarrow$ $\\dfrac{1}{2}*[log_{10}{\\dfrac{2*5*20*50}{4*10*25}}]$<\/p>\n $\\Rightarrow$ $\\dfrac{1}{2}*[log_{10}10]$<\/p>\n $\\Rightarrow$ $\\dfrac{1}{2}$<\/p>\n
\n$\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+1$<\/p>\n
\nLet $2^x = t$
\n=> $(t-5)^2 = 2(t-7\/2)$
\n=> $t^2 + 25 – 10t = 2t – 7$
\n=> $t^2 – 12t + 32 = 0$
\n=> t = 8, 4
\nTherefore, x = 2 or 3, but $2^x$ > 5, so x = 3<\/p>\n
\n$a*log_z y = ab$ => $log_z y = b$
\n$b*log_x z = ab$ => $log_x z = a$
\n$log_y x$ = $log_z y * log_x z$ => $log x\/log y$ = $log y\/log z * log z\/log x$
\n=> $\\frac{log x}{log y} = \\frac{log y}{log x}$
\n=> $(log x)^2 = (log y)^2$
\n=> $log x = log y$ or $log x = -log y$
\nSo, x = y or x = 1\/y
\nSo, ab = 1 or -1
\nOption 5) is not possible<\/p>\n
\nThus, $\\dfrac{\\log {x}}{\\log {10}}$ – $\\dfrac{1}{3}*\\dfrac{\\log {x}}{\\log {10}}$ = $6*\\dfrac{\\log{10}}{\\log{x}}$
\n=> $\\dfrac{2}{3}*\\dfrac{\\log {x}}{\\log {10}}$ = $6*\\dfrac{\\log{10}}{\\log{x}}$
\nThus, => $\\dfrac{1}{9}*(\\log{x})^2 = (\\log{10})^2$
\nThus, $ x = 1000$
\nHence, option D is the correct answer.<\/p>\n
\nHence, 3a = 9 or a = 3<\/p>\n
\n=>$5 +\u00a0\\log_{3}{a}$ = 8
\n=>$\u00a0\\log_{3}{a}$ = 3
\nor $a$ = 27<\/p>\n
\n=>$4a+12+\\log_{2}{b}$ = 125
\nPutting $a$ = 27, we get
\n$\\log_{2}{b}$ = 5
\nor, $b$ = 32<\/p>\n
\nHence, option A is the correct answer.<\/p>\n