{"id":39497,"date":"2019-12-18T17:38:39","date_gmt":"2019-12-18T12:08:39","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=39497"},"modified":"2019-12-18T17:38:39","modified_gmt":"2019-12-18T12:08:39","slug":"logarithm-questions-for-xat-set-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/logarithm-questions-for-xat-set-2-pdf\/","title":{"rendered":"Logarithm Questions for XAT Set-2 PDF"},"content":{"rendered":"

Logarithm Questions for XAT Set-2 PDF<\/strong><\/span><\/h2>\n

Download important Logarithm Questions Set-2 for XAT\u00a0 Set-2 PDF based on previously asked questions in the XAT exam. Practice Logarithm Questions\u00a0 Set-2 PDF for the XAT exam.<\/p>\n

Download Logarithm Questions\u00a0 PDF<\/a><\/p>\n

Get 5 XAT mocks for Rs. 299. Enroll here<\/a><\/p>\n

Question 1:\u00a0<\/b>Let x and y be positive real numbers such that
\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

a)\u00a0150<\/p>\n

b)\u00a025<\/p>\n

c)\u00a0100<\/p>\n

d)\u00a0250<\/p>\n

Question 2:\u00a0<\/b>Sham is trying to solve the expression:
\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

a)\u00a01<\/p>\n

b)\u00a0$\\frac{1}{\\sqrt{2}}$<\/p>\n

c)\u00a00<\/p>\n

d)\u00a0-1<\/p>\n

Question 3:\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

Question 4:\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

Question 5:\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

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Question 6:\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 7:\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 8:\u00a0<\/b>If x is a real number such that $\\log_{3}5= \\log_{5}(2 + x)$, then which of the following is true?<\/p>\n

a)\u00a00 < x < 3<\/p>\n

b)\u00a023 < x < 30<\/p>\n

c)\u00a0x > 30<\/p>\n

d)\u00a03 < x < 23<\/p>\n

Question 9:\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 10:\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

XAT Decision making practice questions<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(A)<\/strong><\/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

2)\u00a0Answer\u00a0(C)<\/strong><\/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

3)\u00a0Answer\u00a0(D)<\/strong><\/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

4)\u00a0Answer\u00a0(D)<\/strong><\/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

5)\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

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6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$\\log_{2}({5+\\log_{3}{a}})=3$
\n=>$5 +\u00a0\\log_{3}{a}$ = 8
\n=>$\u00a0\\log_{3}{a}$ = 3
\nor $a$ = 27<\/p>\n

$\\log_{5}({4a+12+\\log_{2}{b}})=3$
\n=>$4a+12+\\log_{2}{b}$ = 125
\nPutting $a$ = 27, we get
\n$\\log_{2}{b}$ = 5
\nor, $b$ = 32<\/p>\n

So, $a + b$ = 27 + 32 = 59
\nHence, option A is the correct answer.<\/p>\n

7)\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}) $
\nHence, 3a = 9 or a = 3<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$1 < \\log_{3}5 < 2$
\n=> $ 1 < \\log_{5}(2 + x) < 2 $
\n=> $ 5 < 2 + x < 25$
\n=> $ 3 < x < 23$<\/p>\n

9)\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

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$\\log_{10} x – \\log_{10} \\sqrt[3]{x} = 6\\log_{x}10$
\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, => $\\dfrac{1}{9}*(\\log{x})^2 = (\\log{10})^2=1$
\nThus,\u00a0$(\\log{x})^2 = 9$
\nThus $\\log x = 3$ or $-3$
\nThus, $ x = 1000$ or $\\dfrac{1}{1000}$
\nFrom amongst the given options, 1000 is the correct answer.
\nHence, option D is the correct answer.<\/p>\n

XAT Solved Previous papers<\/a><\/p>
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Take XAT Mock Test<\/a><\/p><\/p>\n

We hope this Logarithm Questions Set-2 PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Logarithm Questions for XAT Set-2 PDF Download important Logarithm Questions Set-2 for XAT\u00a0 Set-2 PDF based on previously asked questions in the XAT exam. Practice Logarithm Questions\u00a0 Set-2 PDF for the XAT exam. Question 1:\u00a0Let x and y be positive real numbers such that $\\log_{5}{(x + y)} + \\log_{5}{(x – y)} = 3,$ and $\\log_{2}{y} […]<\/p>\n","protected":false},"author":42,"featured_media":39501,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,3167,125,1352,350,362,366],"tags":[3340,1425],"class_list":{"0":"post-39497","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads-en","9":"category-featured","10":"category-iift-en","11":"category-iift","12":"category-snap","13":"category-xat","14":"tag-logarithm-questions-for-xat","15":"tag-xat"},"better_featured_image":{"id":39501,"alt_text":"Logarithms Questions for XAT Set-2 PDF","caption":"Logarithms Questions for XAT Set-2 PDF\n","description":"Logarithms Questions for XAT Set-2 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