Logarithms is an important topic in the Quant section of the SNAP Exam. You can also download this Free Logarithms Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Logarithms questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Logarithms Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\u00a0<\/b>Find the value of following expression $\\log\\sin 40^\\circ \\log \\sin 41^\\circ — \\log \\sin 99^\\circ\u00a0\\log \\sin 100^\\circ$<\/p>\n a)\u00a0$\\frac{\\sqrt{3ac}+1}{2}$<\/p>\n b)\u00a0$0$<\/p>\n c)\u00a0$1$<\/p>\n d)\u00a0$2$<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n = $\\log\\sin 40^\\circ \\log \\sin 41^\\circ…..\u00a0\\log \\sin 90^\\circ…… \\log \\sin 99^\\circ \\log \\sin 100^\\circ$<\/p>\n = $\\log \\sin 90^\\circ$<\/p>\n = log 1<\/p>\n = 0<\/p>\n Answer is option B.<\/p>\n Question 2:\u00a0<\/b>The value of $\\log_2 x$ which satisfy $6 – 9\\log_{8}\\left(\\frac{4}{x}\\right)^{\\frac{1}{3}} – 8(\\log_{256}x)^{\\frac{2}{3}} – (\\log_2 x^8)^{\\frac{1}{3}} = 0$ is<\/p>\n a)\u00a02<\/p>\n b)\u00a0$\\sqrt{2}$<\/p>\n c)\u00a04<\/p>\n d)\u00a08<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $6-9\\log_8\\left(\\frac{4}{x}\\right)^{\\frac{1}{3}}-8\\left(\\log_{256}x\\right)^{\\frac{2}{3}}-\\left(\\log_{2\\ }x^8\\right)^{\\frac{1}{3}}=0$<\/p>\n $6-\\log_24+\\log_2\\ x-2\\left(\\log_2x\\right)^{\\frac{2}{3}}-2\\left(\\log_2x\\right)^{\\frac{1}{3}}=0$<\/p>\n Let\u00a0$\\left(\\log_2x\\right)^{\\frac{1}{3}}$ be t.<\/p>\n $4+t^3-2t^2-2t=0$<\/p>\n or,\u00a0$\\left(t-2\\right)\\left(t^2-2\\right)=0$<\/p>\n so,\u00a0$t=2$ or<\/p>\n $\\log_2\\ x=8$<\/p>\n Question 3:\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 3)\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 4:\u00a0<\/b>If $\\log_{2}[3+\\log_{3} \\left\\{4+\\log_{4}(x-1) \\right\\}]-2=0$ then 4x equals<\/p>\n 4)\u00a0Answer:\u00a05<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have : Question 5:\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 5)\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 Enroll to SNAP & NMAT 2022 Crash Course<\/a><\/p>\n Question 6:\u00a0<\/b>If $\\log \\left(\\frac{a}{b}\\right) + \\log \\left(\\frac{b}{a}\\right) = \\log(a + b)$, then which of the following statements is CORRECT?<\/p>\n a)\u00a0a – b = 1<\/p>\n b)\u00a0a + b = 1<\/p>\n c)\u00a0a = b<\/p>\n d)\u00a0$a^2 – b^2 = 1$<\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n log(a\/b) + log(b\/a) = log(a+b)<\/p>\n log(a+b) = log(a\/b)(b\/a)<\/p>\n log(a+b)=log 1<\/p>\n a+b=1<\/p>\n Question 7:\u00a0<\/b>If $\\log_4m + \\log_4n = \\log_2(m + n)$ where m and n are positive real numbers, then which of the following must be true?<\/p>\n a)\u00a0$\\frac{1}{m} + \\frac{1}{n} = 1$<\/p>\n b)\u00a0m = n<\/p>\n c)\u00a0$m^2 + n^2 = 1$<\/p>\n d)\u00a0$\\frac{1}{m} + \\frac{1}{n} = 2$<\/p>\n e)\u00a0No values of m and n can satisfy the given equation<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_4mn=\\log_2(m+n)$<\/p>\n $\\sqrt{\\ mn}=(m+n)$<\/p>\n Squarring on both sides<\/p>\n $m^2+n^2+mn\\ =\\ 0$<\/p>\n Since m, n are positive real numbers, no value of m and n satisfy the above equations.<\/p>\n Question 8:\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 8)\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 9:\u00a0<\/b>$\\frac{2\\times4\\times8\\times16}{(\\log_{2}{4})^{2}(\\log_{4}{8})^{3}(\\log_{8}{16})^{4}}$ equals<\/p>\n 9)\u00a0Answer:\u00a024<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\left(2\\cdot4\\cdot8\\cdot16\\right)}{\\left(\\log_22^2\\right)^2\\cdot\\left(\\log_{2^2}2^3\\right)^3\\cdot\\left(\\log_{2^3}2^4\\right)^4}\\cdot$<\/p>\n =\u00a0$\\frac{2^{10}}{4\\cdot\\left(\\frac{3}{2}\\right)^3\\cdot\\left(\\frac{4}{3}\\right)^4}=24$<\/p>\n Question 10:\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 10)\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 Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 11:\u00a0<\/b>If $\\log_{4}{5}=(\\log_{4}{y})(\\log_{6}{\\sqrt{5}})$, then y equals<\/p>\n 11)\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 12:\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 12)\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 13:\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 13)\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 14:\u00a0<\/b>Sham is trying to solve the expression: a)\u00a01<\/p>\n b)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n c)\u00a00<\/p>\n d)\u00a0-1<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log \\tan 1^\\circ + \\log \\tan 2^\\circ + \\log \\tan 3^\\circ + …….. + \\log \\tan 89^\\circ$.<\/p>\n =$\\log \\tan 1^\\circ + \\log \\tan 89^\\circ + \\log \\tan 2^\\circ + \\log \\tan 88^\\circ …….. + \\log \\tan 45^\\circ$.<\/p>\n =$\\log\\ \\left(\\tan\\ 1^0\\cdot\\tan\\ 89^0\\right)\\times\\log\\ \\left(\\tan\\ 2^0\\cdot\\tan\\ 88^0\\right)\\ ………………………\\log\\ \\left(\\tan\\ 45^0\\right)$<\/p>\n tan $45^0$ = 1<\/p>\n $\\log\\ \\left(\\tan\\ 45^0\\right)\\ =\\ 0$<\/p>\n $\\therefore$\u00a0$\\log \\tan 1^\\circ + \\log \\tan 2^\\circ + \\log \\tan 3^\\circ + …….. + \\log \\tan 89^\\circ$ = 0<\/p>\n Question 15:\u00a0<\/b>If $\\log_{10}{11} = a$ then $\\log_{10}{\\left(\\frac{1}{110}\\right)}$ is equal to<\/p>\n a)\u00a0$-a$<\/p>\n b)\u00a0$(1 + a)^{-1}$<\/p>\n c)\u00a0$\\frac{1}{10 a}$<\/p>\n d)\u00a0$-(a + 1)$<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{10}{\\left(\\frac{1}{110}\\right)}$<\/p>\n $\\log_a\\left(\\ \\frac{\\ x}{y}\\right)\\ =\\ \\log_ax-\\log_ay$<\/p>\n $\\log_{10}{\\left(\\frac{1}{110}\\right)}$ =\u00a0$=\\ \\log_{10}1-\\log_{10}110$<\/p>\n = 0$-\\log_{10}110$<\/p>\n =$-\\log_{10}11\\times\\ 10$<\/p>\n =$-\\left(\\log_{10}11+\\log_{10}10\\right)$<\/p>\n = -(a+1)<\/p>\n D is the correct answer.<\/p>\n Question 16:\u00a0<\/b>Find the value of $\\log_{10}{10} + \\log_{10}{10^2} + ….. + \\log_{10}{10^n}$<\/p>\n a)\u00a0$n^{2} + 1$<\/p>\n b)\u00a0$n^{2} – 1$<\/p>\n c)\u00a0$\\frac{(n^{2} + n)}{2}.\\frac{n(n + 1)}{3}$<\/p>\n d)\u00a0$\\frac{(n^{2} + n)}{2}$<\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{10}{10} + \\log_{10}{10^2} + ….. + \\log_{10}{10^n}$<\/p>\n Since\u00a0$\\log_aa\\ $ = 1<\/p>\n $\\log_{10}{10} + \\log_{10}{10^2} + ….. + \\log_{10}{10^n}$ = 1+2+….n<\/p>\n =$\\ \\frac{\\ n\\left(n+1\\right)}{2}$<\/p>\n =$\\frac{(n^{2} + n)}{2}$<\/p>\n D is the correct answer.<\/p>\n Question 17:\u00a0<\/b>what is the value of $\\frac{\\log_{27}{9} \\times \\log_{16}{64}}{\\log_{4}{\\sqrt2}}$?<\/p>\n a)\u00a0$\\frac{1}{6}$<\/p>\n b)\u00a0$\\frac{1}{4}$<\/p>\n c)\u00a08<\/p>\n d)\u00a04<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\log_{27}{9} \\times \\log_{16}{64}}{\\log_{4}{\\sqrt2}}$?<\/p>\n =$\\frac{\\ \\log_{3^3}3^2\\times\\ \\log_{2^4}2^6}{\\log_{\\left(\\sqrt{\\ 2}\\right)^4}\\sqrt{\\ 2}}$<\/p>\n =$\\frac{\\ \\ \\frac{\\ 2}{3}\\times\\ \\frac{\\ 6}{4}}{\\ \\frac{\\ 1}{4}}$<\/p>\n =4<\/p>\n D is the correct answer.<\/p>\n Question 18:\u00a0<\/b>What is the value of x in the following expression? a)\u00a01<\/p>\n b)\u00a00<\/p>\n c)\u00a0-1<\/p>\n d)\u00a03<\/p>\n 18)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The given equation can be written as<\/p>\n $\\log\\left(10\\right)^{x\\ }\\ +\\ \\log\\left(1+2^x\\right)=\\log\\left(5\\right)^x+\\log6$<\/p>\n $\\log\\left(10\\right)^{x\\ }\\left(1+2^x\\right)=\\log\\left(5\\right)^x\\cdot6$\u00a0 \u00a0 (\u00a0 since logA + logB=logAB)<\/p>\n $\\log\\ \\frac{\\left(2^x\\cdot5^x\\right)\\left(1+2^x\\right)}{5^x\\cdot6}=0$\u00a0 \u00a0\u00a0( since logA – logB=logA\/B)<\/p>\n $\\frac{\\left(2^x\\ +2^{2x\\ }\\right)}{6}=10^0$\u00a0 ($Since\\ \\log_aN\\ =x\\ \\ =>N=a^x$)<\/p>\n $2^{^x}+2^{2x}=6$<\/p>\n The above\u00a0\u00a0equation is satisfied only when x=1<\/p>\n Question 19:\u00a0<\/b>Find the value of $\\log_{3^2}{5^4} \\times \\log_{5^2}{3^4}$<\/p>\n a)\u00a05<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a02<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log_{b^n}\\left(a^m\\right)\\ =\\frac{m}{n}\\log_ba\\ =\\frac{m}{n}\\cdot\\frac{\\log\\left(a\\right)}{\\log\\left(b\\right)}$<\/p>\n So given equation becomes\u00a0$\\frac{4}{2}\\cdot\\frac{4}{2}\\cdot\\frac{\\log\\left(3\\right)}{\\log\\left(2\\right)}\\cdot\\frac{\\log\\left(2\\right)}{\\log\\left(3\\right)}$\u00a0 = 4<\/p>\n Question 20:\u00a0<\/b>$\\log_{5}{25} + \\log_{2} (\\log_{3}{81})$ is<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\log\\left(a^m\\right)\\ =\\ m\\log\\left(a\\right)\\ and\\ \\ \\log_aa$ = 1<\/p>\n $\\log_55^2\\ +\\ \\log_2\\left(\\log_33^4\\right)$<\/p>\n 2 +\u00a0$\\ \\log_24$<\/p>\n 2+\u00a0$\\ \\log_22^2$<\/p>\n 4<\/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
\n$\\log \\tan 1^\\circ + \\log \\tan 2^\\circ + \\log \\tan 3^\\circ\u00a0+ …….. +\u00a0\\log \\tan 89^\\circ$.
\nThe correct answer would be?<\/p>\n
\n$x + \\log_{10} (1 + 2^x) = x \\log_{10} 5 + \\log_{10} 6$<\/p>\n