{"id":207138,"date":"2021-12-22T16:34:13","date_gmt":"2021-12-22T11:04:13","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=207138"},"modified":"2021-12-22T16:34:13","modified_gmt":"2021-12-22T11:04:13","slug":"logarithms-questions-for-xat-2022-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/logarithms-questions-for-xat-2022-pdf\/","title":{"rendered":"Logarithms Questions for XAT 2022 &#8211; Download PDF"},"content":{"rendered":"<h1>Logarithms Questions for XAT 2022 &#8211; Download PDF<\/h1>\n<p>Download Logarithms Questions for XAT PDF \u2013 XAT Logarithms questions pdf by Cracku. Top 10 very Important Logarithms Questions for XAT based on asked questions in previous exam papers.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/downloads\/14011\" target=\"_blank\" class=\"btn btn-danger  download\">Download Logarithms Questions for XAT <\/a><\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/pay\/aDOMI\" target=\"_blank\" class=\"btn btn-info \">Get 5 XAT mocks at just Rs.299<\/a><\/p>\n<p><b>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<p>a)\u00a0$2&lt;a&lt;3$<\/p>\n<p>b)\u00a0$3&lt;a&lt;4$<\/p>\n<p>c)\u00a0$4&lt;a&lt;5$<\/p>\n<p>d)\u00a0$a&gt;5$<\/p>\n<p><b>Question 2:\u00a0<\/b>If $\\log_{2}[3+\\log_{3} \\left\\{4+\\log_{4}(x-1) \\right\\}]-2=0$ then 4x equals<\/p>\n<p><b>Question 3:\u00a0<\/b>If $5 &#8211; \\log_{10}\\sqrt{1 + x} + 4 \\log_{10} \\sqrt{1 &#8211; x} = \\log_{10} \\frac{1}{\\sqrt{1 &#8211; x^2}}$, then 100x equals<\/p>\n<p><b>Question 4:\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<p>a)\u00a0$\\frac{1}{m} + \\frac{1}{n} = 1$<\/p>\n<p>b)\u00a0m = n<\/p>\n<p>c)\u00a0$m^2 + n^2 = 1$<\/p>\n<p>d)\u00a0$\\frac{1}{m} + \\frac{1}{n} = 2$<\/p>\n<p>e)\u00a0No values of m and n can satisfy the given equation<\/p>\n<p><b>Question 5:\u00a0<\/b>The value of $\\log_{a}({\\frac{a}{b}})+\\log_{b}({\\frac{b}{a}})$, for $1&lt;a\\leq b$ cannot be equal to<\/p>\n<p>a)\u00a00<\/p>\n<p>b)\u00a0-1<\/p>\n<p>c)\u00a01<\/p>\n<p>d)\u00a0-0.5<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/xat-crash-course\" target=\"_blank\" class=\"btn btn-warning \">XAT 2022 Crash Course<\/a><\/p>\n<p><b>Question 6:\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<p>a)\u00a0$\\frac{2}{A+B-3}$<\/p>\n<p>b)\u00a0$\\frac{2}{A+B}-3$<\/p>\n<p>c)\u00a0$\\frac{A+B}{2}-3$<\/p>\n<p>d)\u00a0$\\frac{A+B-3}{2}$<\/p>\n<p><b>Question 7:\u00a0<\/b>If $\\log_{4}{5}=(\\log_{4}{y})(\\log_{6}{\\sqrt{5}})$, then y equals<\/p>\n<p><b>Question 8:\u00a0<\/b>If $\\log_{10}{11} = a$ then $\\log_{10}{\\left(\\frac{1}{110}\\right)}$ is equal to<\/p>\n<p>a)\u00a0$-a$<\/p>\n<p>b)\u00a0$(1 + a)^{-1}$<\/p>\n<p>c)\u00a0$\\frac{1}{10 a}$<\/p>\n<p>d)\u00a0$-(a + 1)$<\/p>\n<p><b>Question 9:\u00a0<\/b>Find the value of $\\log_{10}{10} + \\log_{10}{10^2} + &#8230;.. + \\log_{10}{10^n}$<\/p>\n<p>a)\u00a0$n^{2} + 1$<\/p>\n<p>b)\u00a0$n^{2} &#8211; 1$<\/p>\n<p>c)\u00a0$\\frac{(n^{2} + n)}{2}.\\frac{n(n + 1)}{3}$<\/p>\n<p>d)\u00a0$\\frac{(n^{2} + n)}{2}$<\/p>\n<p><b>Question 10:\u00a0<\/b>what is the value of $\\frac{\\log_{27}{9} \\times \\log_{16}{64}}{\\log_{4}{\\sqrt2}}$?<\/p>\n<p>a)\u00a0$\\frac{1}{6}$<\/p>\n<p>b)\u00a0$\\frac{1}{4}$<\/p>\n<p>c)\u00a08<\/p>\n<p>d)\u00a04<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/xat-mock-test\" target=\"_blank\" class=\"btn btn-danger \">Take XAT 2022 mocks<\/a><\/p>\n<p><span style=\"text-decoration: underline;\"><strong>Answers &amp; Solutions:<\/strong><\/span><\/p>\n<p><strong>1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>We have :$\\frac{\\log_{15}{a}+\\log_{32}{a}}{(\\log_{15}{a})(\\log_{32}{a})}=4$<br \/>\nWe get $\\frac{\\left(\\frac{\\log a}{\\log\\ 15}+\\frac{\\log a}{\\log32}\\right)}{\\frac{\\log a}{\\log\\ 15}\\times\\ \\frac{\\log a}{\\log32}\\ \\ }=4$<br \/>\nwe get $\\log a\\left(\\log32\\ +\\log\\ 15\\right)=4\\left(\\log\\ a\\right)^2$<br \/>\nwe get $\\left(\\log32\\ +\\log\\ 15\\right)=4\\log a$<br \/>\n=$\\log480=\\log a^4$<br \/>\n=$a^4\\ =480$<br \/>\nso we can say a is between 4 and 5 .<\/p>\n<p><b>2)\u00a0Answer:\u00a05<\/b><\/p>\n<p>We have :<br \/>\n$\\log_2\\left\\{3+\\log_3\\left\\{4+\\log_4\\left(x-1\\right)\\right\\}\\right\\}=2$<br \/>\nwe get\u00a0$3+\\log_3\\left\\{4+\\log_4\\left(x-1\\right)\\right\\}=4$<br \/>\nwe get\u00a0$\\log_3\\left(4+\\log_4\\left(x-1\\right)\\ =\\ 1\\right)$<br \/>\nwe get\u00a0$4+\\log_4\\left(x-1\\right)\\ =\\ 3$<br \/>\n$\\log_4\\left(x-1\\right)\\ =\\ -1$<br \/>\nx-1 = 4^-1<br \/>\nx =\u00a0$\\frac{1}{4}+1=\\frac{5}{4}$<br \/>\n4x = 5<\/p>\n<p><b>3)\u00a0Answer:\u00a099<\/b><\/p>\n<p>$5 &#8211; \\log_{10}\\sqrt{1 + x} + 4 \\log_{10} \\sqrt{1 &#8211; x} = \\log_{10} \\frac{1}{\\sqrt{1 &#8211; x^2}}$<\/p>\n<p>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<p>$5-\\log_{10}\\sqrt{1+x}+4\\log_{10}\\sqrt{1-x}=\\left(-1\\right)\\log_{10}\\left(\\sqrt{1+x}\\right)+\\left(-1\\right)\\log_{10}\\left(\\sqrt{1-x}\\right)$<\/p>\n<p>$5=-\\log_{10}\\sqrt{1+x}+\\log_{10}\\sqrt{1+x}-\\log_{10}\\sqrt{1-x}-4\\log_{10}\\sqrt{1-x}$<\/p>\n<p>$5=-5\\log_{10}\\sqrt{1-x}$<\/p>\n<p>$\\sqrt{1-x}=\\frac{1}{10}$<\/p>\n<p>Squaring both sides:\u00a0$\\left(\\sqrt{1-x}\\right)^2=\\frac{1}{100}$<\/p>\n<p>$\\therefore\\ $\u00a0$x=1-\\frac{1}{100}=\\frac{99}{100}$<\/p>\n<p>Hence,\u00a0$100\\ x\\ =100\\times\\ \\frac{99}{100}=99$<\/p>\n<p><strong>4)\u00a0Answer\u00a0(E)<\/strong><\/p>\n<p>$\\log_4mn=\\log_2(m+n)$<\/p>\n<p>$\\sqrt{\\ mn}=(m+n)$<\/p>\n<p>Squarring on both sides<\/p>\n<p>$m^2+n^2+mn\\ =\\ 0$<\/p>\n<p>Since m, n are positive real numbers, no value of m and n satisfy the above equations.<\/p>\n<p><strong>5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n<p>On expanding the expression we get\u00a0$1-\\log_ab+1-\\log_ba$<\/p>\n<p>$or\\ 2-\\left(\\log_ab+\\frac{1}{\\log_ba}\\right)$<\/p>\n<p>Now applying the property of AM&gt;=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<p><strong>6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n<p>$\\log_a30=A\\ or\\ \\log_a5+\\log_a2+\\log_a3=A$&#8230;&#8230;&#8230;..(1)<\/p>\n<p>$\\log_a\\left(\\frac{5}{3}\\right)=-B\\ or\\ \\log_a3-\\log_a5=B$&#8230;&#8230;&#8230;&#8230;.(2)<\/p>\n<p>and finally $\\log_a2=3$<\/p>\n<p>Substituting this in (1) we get $\\log_a5+\\log_a3=A-3$<\/p>\n<p>Now we have two equations in two variables (1) and (2) . On solving we get<\/p>\n<p>$\\log_a3=\\frac{\\left(A+B-3\\right)}{2\\ }or\\ \\log_3a=\\frac{2}{A+B-3}$<\/p>\n<p><b>7)\u00a0Answer:\u00a036<\/b><\/p>\n<p>$\\frac{\\log\\ 5}{2\\log2}\\ =\\frac{\\log\\ y}{2\\log2}\\cdot\\frac{\\log\\ 5}{2\\log6}$<\/p>\n<p>$\\log\\ 36\\ =\\ \\log\\ y;\\ \\therefore\\ y\\ =36$<\/p>\n<p><strong>8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>$\\log_{10}{\\left(\\frac{1}{110}\\right)}$<\/p>\n<p>$\\log_a\\left(\\ \\frac{\\ x}{y}\\right)\\ =\\ \\log_ax-\\log_ay$<\/p>\n<p>$\\log_{10}{\\left(\\frac{1}{110}\\right)}$ =\u00a0$=\\ \\log_{10}1-\\log_{10}110$<\/p>\n<p>= 0$-\\log_{10}110$<\/p>\n<p>=$-\\log_{10}11\\times\\ 10$<\/p>\n<p>=$-\\left(\\log_{10}11+\\log_{10}10\\right)$<\/p>\n<p>= -(a+1)<\/p>\n<p>D is the correct answer.<\/p>\n<p><strong>9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>$\\log_{10}{10} + \\log_{10}{10^2} + &#8230;.. + \\log_{10}{10^n}$<\/p>\n<p>Since\u00a0$\\log_aa\\ $ = 1<\/p>\n<p>$\\log_{10}{10} + \\log_{10}{10^2} + &#8230;.. + \\log_{10}{10^n}$ = 1+2+&#8230;.n<\/p>\n<p>=$\\ \\frac{\\ n\\left(n+1\\right)}{2}$<\/p>\n<p>=$\\frac{(n^{2} + n)}{2}$<\/p>\n<p>D is the correct answer.<\/p>\n<p><strong>10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n<p>$\\frac{\\log_{27}{9} \\times \\log_{16}{64}}{\\log_{4}{\\sqrt2}}$?<\/p>\n<p>=$\\frac{\\ \\log_{3^3}3^2\\times\\ \\log_{2^4}2^6}{\\log_{\\left(\\sqrt{\\ 2}\\right)^4}\\sqrt{\\ 2}}$<\/p>\n<p>=$\\frac{\\ \\ \\frac{\\ 2}{3}\\times\\ \\frac{\\ 6}{4}}{\\ \\frac{\\ 1}{4}}$<\/p>\n<p>=4<\/p>\n<p>D is the correct answer.<\/p>\n<p class=\"text-center\"><a href=\"https:\/\/cracku.in\/xat-previous-papers\" target=\"_blank\" class=\"btn btn-info \">Download XAT previous papers [PDF]<\/a><\/p>\n<p>We hope these Logarithms questions for XAT pdf for the XAT exam will be highly useful for your preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Logarithms Questions for XAT 2022 &#8211; Download PDF Download Logarithms Questions for XAT PDF \u2013 XAT Logarithms questions pdf by Cracku. Top 10 very Important Logarithms Questions for XAT based on asked questions in previous exam papers. Question 1:\u00a0For 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 a)\u00a0$2&lt;a&lt;3$ b)\u00a0$3&lt;a&lt;4$ c)\u00a0$4&lt;a&lt;5$ [&hellip;]<\/p>\n","protected":false},"author":32,"featured_media":207169,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,3167,125,366],"tags":[3150,3142,1425,4987],"class_list":{"0":"post-207138","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-downloads-en","9":"category-featured","10":"category-xat","11":"tag-logarithms","12":"tag-logarithms-questions-for-xat","13":"tag-xat","14":"tag-xat-2022"},"better_featured_image":{"id":207169,"alt_text":"Logarithms Questions for XAT","caption":"Logarithms Questions for XAT","description":"Logarithms Questions 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