The logarithm is one of the most important topics in the CAT Quantitative Ability Section. You can check out these Logarithm questions in<\/span> CAT Previous year papers<\/a>. <\/strong><\/span>If you want to learn the basics, you can watch these videos on Logarithm basics<\/strong><\/span>. This article will look into some important Logarithm Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Logarithm Most Important Questions PDF below, which is completely Free.<\/p>\n Download Logarithm Questions for CAT<\/a><\/p>\n CAT Online Coaching<\/a><\/p>\n Question 1:\u00a0<\/b>For a real number a, if $\\frac{\\log_{15}{a}+\\log_{32}{a}}{(\\log_{15}{a})(\\log_{32}{a})}=4$ then a must lie in the range<\/p>\n a)\u00a0$2<a<3$<\/p>\n b)\u00a0$3<a<4$<\/p>\n c)\u00a0$4<a<5$<\/p>\n d)\u00a0$a>5$<\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have :$\\frac{\\log_{15}{a}+\\log_{32}{a}}{(\\log_{15}{a})(\\log_{32}{a})}=4$ Question 2:\u00a0<\/b>If $\\log_{2}[3+\\log_{3} \\left\\{4+\\log_{4}(x-1) \\right\\}]-2=0$ then 4x equals<\/p>\n 2)\u00a0Answer:\u00a05<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have : Question 3:\u00a0<\/b>If $5 – \\log_{10}\\sqrt{1 + x} + 4 \\log_{10} \\sqrt{1 – x} = \\log_{10} \\frac{1}{\\sqrt{1 – x^2}}$, then 100x equals<\/p>\n 3)\u00a0Answer:\u00a099<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $5 – \\log_{10}\\sqrt{1 + x} + 4 \\log_{10} \\sqrt{1 – x} = \\log_{10} \\frac{1}{\\sqrt{1 – x^2}}$<\/p>\n We can re-write the equation as:\u00a0$5-\\log_{10}\\sqrt{1+x}+4\\log_{10}\\sqrt{1-x}=\\log_{10}\\left(\\sqrt{1+x}\\times\\ \\sqrt{1-x}\\right)^{-1}$<\/p>\n $5-\\log_{10}\\sqrt{1+x}+4\\log_{10}\\sqrt{1-x}=\\left(-1\\right)\\log_{10}\\left(\\sqrt{1+x}\\right)+\\left(-1\\right)\\log_{10}\\left(\\sqrt{1-x}\\right)$<\/p>\n $5=-\\log_{10}\\sqrt{1+x}+\\log_{10}\\sqrt{1+x}-\\log_{10}\\sqrt{1-x}-4\\log_{10}\\sqrt{1-x}$<\/p>\n $5=-5\\log_{10}\\sqrt{1-x}$<\/p>\n $\\sqrt{1-x}=\\frac{1}{10}$<\/p>\n Squaring both sides:\u00a0$\\left(\\sqrt{1-x}\\right)^2=\\frac{1}{100}$<\/p>\n $\\therefore\\ $\u00a0$x=1-\\frac{1}{100}=\\frac{99}{100}$<\/p>\n Hence,\u00a0$100\\ x\\ =100\\times\\ \\frac{99}{100}=99$<\/p>\n Question 4:\u00a0<\/b>The value of $\\log_{a}({\\frac{a}{b}})+\\log_{b}({\\frac{b}{a}})$, for $1<a\\leq b$ cannot be equal to<\/p>\n a)\u00a00<\/p>\n b)\u00a0-1<\/p>\n c)\u00a01<\/p>\n d)\u00a0-0.5<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n On expanding the expression we get\u00a0$1-\\log_ab+1-\\log_ba$<\/p>\n $or\\ 2-\\left(\\log_ab+\\frac{1}{\\log_ba}\\right)$<\/p>\n Now applying the property of AM>=GM, we get that\u00a0\u00a0$\\frac{\\left(\\log_ab+\\frac{1}{\\log_ba}\\right)}{2}\\ge1\\ or\\ \\left(\\log_ab+\\frac{1}{\\log_ba}\\right)\\ge2$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.<\/p>\n Question 5:\u00a0<\/b>If $\\log_{a}{30}=A,\\log_{a}({\\frac{5}{3}})=-B$ and $\\log_2{a}=\\frac{1}{3}$, then $\\log_3{a}$ equals<\/p>\n a)\u00a0$\\frac{2}{A+B-3}$<\/p>\n b)\u00a0$\\frac{2}{A+B}-3$<\/p>\n c)\u00a0$\\frac{A+B}{2}-3$<\/p>\n d)\u00a0$\\frac{A+B-3}{2}$<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\log_a30=A\\ or\\ \\log_a5+\\log_a2+\\log_a3=A$………..(1)<\/p>\n $\\log_a\\left(\\frac{5}{3}\\right)=-B\\ or\\ \\log_a3-\\log_a5=B$………….(2)<\/p>\n and finally $\\log_a2=3$<\/p>\n Substituting this in (1) we get $\\log_a5+\\log_a3=A-3$<\/p>\n Now we have two equations in two variables (1) and (2) . On solving we get<\/p>\n $\\log_a3=\\frac{\\left(A+B-3\\right)}{2\\ }or\\ \\log_3a=\\frac{2}{A+B-3}$<\/p>\n Checkout: CAT Free Practice Questions and Videos<\/p>\n Question 6:\u00a0<\/b>If $\\log_{4}{5}=(\\log_{4}{y})(\\log_{6}{\\sqrt{5}})$, then y equals<\/p>\n 6)\u00a0Answer:\u00a036<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\log\\ 5}{2\\log2}\\ =\\frac{\\log\\ y}{2\\log2}\\cdot\\frac{\\log\\ 5}{2\\log6}$<\/p>\n $\\log\\ 36\\ =\\ \\log\\ y;\\ \\therefore\\ y\\ =36$<\/p>\n Question 7:\u00a0<\/b>If Y is a negative number such that $2^{Y^2({\\log_{3}{5})}}=5^{\\log_{2}{3}}$, then Y equals to:<\/p>\n a)\u00a0$\\log_{2}(\\frac{1}{5})$<\/p>\n b)\u00a0$\\log_{2}(\\frac{1}{3})$<\/p>\n c)\u00a0$-\\log_{2}(\\frac{1}{5})$<\/p>\n d)\u00a0$-\\log_{2}(\\frac{1}{3})$<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $2^{Y^2({\\log_{3}{5})}}=5^{Y^2(\\log_3 2)}$<\/p>\n Given,\u00a0$5^{Y^2\\left(\\log_32\\right)}=5^{\\left(\\log_23\\right)}$<\/p>\n =>\u00a0$Y^2\\left(\\log_32\\right)=\\left(\\log_23\\right)=>Y^2=\\left(\\log_23\\right)^2$<\/p>\n =>$Y=\\left(-\\log_23\\right)^{\\ }or\\ \\left(\\log_23\\right)$<\/p>\n since Y is a negative number, Y=$\\left(-\\log_23\\right)=\\left(\\log_2\\frac{1}{3}\\right)$<\/p>\n Question 8:\u00a0<\/b>Let x and y be positive real numbers such that a)\u00a0150<\/p>\n b)\u00a025<\/p>\n c)\u00a0100<\/p>\n d)\u00a0250<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have,\u00a0$\\log_{5}{(x + y)} + \\log_{5}{(x – y)} = 3$<\/p>\n =>\u00a0$x^2-y^2=125$……(1)<\/p>\n $\\log_{2}{y} – \\log_{2}{x} = 1 – \\log_{2}{3}$<\/p>\n =>$\\ \\frac{\\ y}{x}$ =\u00a0$\\ \\frac{\\ 2}{3}$<\/p>\n => 2x=3y\u00a0 \u00a0=> x=$\\ \\frac{\\ 3y}{2}$<\/p>\n On substituting the value of x in 1, we get<\/p>\n $\\ \\frac{\\ 5x^2}{4}$=125<\/p>\n =>y=10, x=15<\/p>\n Hence xy=150<\/p>\n Question 9:\u00a0<\/b>If p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$, then the value of $log_{s}{(pqr)}$ is equal to<\/p>\n a)\u00a0$\\frac{47}{10}$<\/p>\n b)\u00a0$\\frac{24}{5}$<\/p>\n c)\u00a0$\\frac{16}{5}$<\/p>\n d)\u00a0$1$<\/p>\n
\nWe get $\\frac{\\left(\\frac{\\log a}{\\log\\ 15}+\\frac{\\log a}{\\log32}\\right)}{\\frac{\\log a}{\\log\\ 15}\\times\\ \\frac{\\log a}{\\log32}\\ \\ }=4$
\nwe get $\\log a\\left(\\log32\\ +\\log\\ 15\\right)=4\\left(\\log\\ a\\right)^2$
\nwe get $\\left(\\log32\\ +\\log\\ 15\\right)=4\\log a$
\n=$\\log480=\\log a^4$
\n=$a^4\\ =480$
\nso we can say a is between 4 and 5 .<\/p>\n
\n$\\log_2\\left\\{3+\\log_3\\left\\{4+\\log_4\\left(x-1\\right)\\right\\}\\right\\}=2$
\nwe get\u00a0$3+\\log_3\\left\\{4+\\log_4\\left(x-1\\right)\\right\\}=4$
\nwe get\u00a0$\\log_3\\left(4+\\log_4\\left(x-1\\right)\\ =\\ 1\\right)$
\nwe get\u00a0$4+\\log_4\\left(x-1\\right)\\ =\\ 3$
\n$\\log_4\\left(x-1\\right)\\ =\\ -1$
\nx-1 = 4^-1
\nx =\u00a0$\\frac{1}{4}+1=\\frac{5}{4}$
\n4x = 5<\/p>\n
\n$\\log_{5}{(x + y)} + \\log_{5}{(x – y)} = 3,$ and $\\log_{2}{y} – \\log_{2}{x} = 1 – \\log_{2}{3}$. Then $xy$ equals<\/p>\n