{"id":38572,"date":"2019-12-02T11:58:21","date_gmt":"2019-12-02T06:28:21","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=38572"},"modified":"2019-12-02T11:58:21","modified_gmt":"2019-12-02T06:28:21","slug":"logarithms-questions-for-xat-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/logarithms-questions-for-xat-pdf\/","title":{"rendered":"Logarithms Questions for XAT (PDF)"},"content":{"rendered":"

Logarithms Questions for XAT (PDF)<\/strong><\/span><\/h2>\n

Download important Logarithms Questions for XAT PDF based on previously asked questions in XAT exam. Practice Logarithms Questions PDF for XAT exam.<\/p>\n

Download Logarithms Questions for XAT PDF<\/a><\/p>\n

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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>Let $u = ({\\log_2 x})^2 – 6 {\\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has<\/p>\n

a)\u00a0no solution for x<\/p>\n

b)\u00a0exactly one solution for x<\/p>\n

c)\u00a0exactly two distinct solutions for x<\/p>\n

d)\u00a0exactly three distinct solutions for x<\/p>\n

Question 3:\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 4:\u00a0<\/b>If x >= y and y > 1, then the value of the expression $log_x (x\/y) + log_y (y\/x)$ can never be<\/p>\n

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

b)\u00a0-0.5<\/p>\n

c)\u00a00<\/p>\n

d)\u00a01<\/p>\n

Question 5:\u00a0<\/b>If $\\log_{2}{\\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of \u2018x\u2019?<\/p>\n

a)\u00a03<\/p>\n

b)\u00a05<\/p>\n

c)\u00a04<\/p>\n

d)\u00a0None of these<\/p>\n

XAT Solved Previous papers<\/a><\/p>
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Question 6:\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 7:\u00a0<\/b>What is the value of $\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+1$<\/p>\n

a)\u00a0-5\/3<\/p>\n

b)\u00a02<\/p>\n

c)\u00a05\/3<\/p>\n

d)\u00a01<\/p>\n

Question 8:\u00a0<\/b>Find the value of x from the following equation:
\n$\\log_{10}{3}+\\log_{10}(4x+1)=\\log_{10}(x+1)+1$<\/p>\n

a)\u00a02\/7<\/p>\n

b)\u00a07\/2<\/p>\n

c)\u00a09\/2<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 9:\u00a0<\/b>If $\\log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression, then x is equal to<\/p>\n

a)\u00a0$\\frac{8}{3}$<\/p>\n

b)\u00a0$\\log_{3}{8}$<\/p>\n

c)\u00a0$\\log_{2}{3}$<\/p>\n

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

$2 log (2^x – 5) = log 2 + log (2^x – 7\/2)$
\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

2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

$x^u = 256$<\/p>\n

Taking log to the base 2 on both the sides,<\/p>\n

$u * \\log_{2}{x} = \\log_{2}{256}$<\/p>\n

=>$[({\\log_2 x})^2 – 6 {\\log_2 x} + 12] * \\log_{2}{x} = 8$<\/p>\n

$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$<\/p>\n

Let $log_2 x = t$<\/p>\n

$t^3 – 6t^2 +12t – 8 = 0$<\/p>\n

$(t-2)^3 = 0$<\/p>\n

Therefore, $log_2 x = 2$<\/p>\n

=> $x = 4$ is the only solution<\/p>\n

Hence, option B is the correct answer.<\/p>\n

3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n

$log_y x = ab$
\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

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

$log_x (x\/y) + log_y (y\/x)$ = $1 – log_x (y) + 1 – log_y (x)$
\n= $2 – (log_x y + 1\/log_x y)$ <= 0 (Since $log_x y + 1\/log_x y$ >= 2)
\nSo, the value of the expression cannot be 1.<\/p>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\log_{2}{\\log_{7}{(x^2 – x+37)}}$ = 1<\/p>\n

$\\log_{7}{(x^2 – x+37)}$ = $2$<\/p>\n

$(x^2 – x+37)$ = $7^{2}$<\/p>\n

Given eq. can be reduced to $x^2 – x + 37 = 49$<\/p>\n

So x can be either -3 or 4.<\/p>\n

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6)\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

7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+\\log_{4}{4}$<\/p>\n

i.e. $\\log_{4}\\log_{4}4^{a-b}=\\log_{4}((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. $\\log_{4}4^{a-b}=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. (a-b)*$\\log_{4}4=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. a-b = 4a+4b-8$\\sqrt{ab}$<\/p>\n

i.e. 3a + 5b – 8$\\sqrt{ab}$ = 0<\/p>\n

i.e. $3\\sqrt\\frac{a}{b}^2$ – 8$\\sqrt\\frac{a}{b}$+5 = 0<\/p>\n

put\u00a0$\\sqrt\\frac{a}{b}$ = t<\/p>\n

therefore 3$t^2$ – 8t + 5 = 0<\/p>\n

solving we get t = 1 or t = $\\frac{5}{3}$<\/p>\n

i.e.\u00a0$\\sqrt\\frac{a}{b}$ = 1 or\u00a0$\\frac{5}{3}$<\/p>\n

but if\u00a0$\\sqrt\\frac{a}{b}$ = 1 then a=b then $\\log_{4}(\\sqrt{a}-\\sqrt{b})$ will become indefinite<\/p>\n

Therefore\u00a0\u00a0$\\sqrt\\frac{a}{b}$ =\u00a0$\\frac{5}{3}$<\/p>\n

Therefore our answer is option ‘C’<\/p>\n

8)\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

9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

If $log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression
\nThen, $2*log(3^{x} – 2) = log{3}+log (3^{x}+ 4)$
\nThus, $log{(3^{x} – 2)^2} = log{3(3^x+4)}$
\nThus, $(3^{x} – 2)^2 = 3(3^x+4)$
\n=> $3^{2x} – 4*3^x +4 = 3*3^x + 12$
\n=> $3^{2x} – 7*3^x – 8 = 0$
\n=> $(3^x+1)*(3^x-8) = 0$
\nBut $3^x+1 \\neq 0$
\nThus, $3^x = 8$
\nHence, $x = log_{3}{8}$
\nHence, option B is the correct answer.<\/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

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We hope this Logarithms\u00a0 Questions PDF for XAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

Logarithms Questions for XAT (PDF) Download important Logarithms Questions for XAT PDF based on previously asked questions in XAT exam. Practice Logarithms Questions PDF for XAT exam. Question 1:\u00a0If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7\/2)$ are in arithmetic progression, then the value of x is equal to a)\u00a05 b)\u00a04 c)\u00a02 d)\u00a03 […]<\/p>\n","protected":false},"author":42,"featured_media":38577,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,169,125,350,1352,362,366],"tags":[3150,3142,1346],"class_list":{"0":"post-38572","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads","9":"category-featured","10":"category-iift","11":"category-iift-en","12":"category-snap","13":"category-xat","14":"tag-logarithms","15":"tag-logarithms-questions-for-xat","16":"tag-xat-2019"},"better_featured_image":{"id":38577,"alt_text":"Logarithms Questions for XAT","caption":"Logarithms Questions for XAT\n","description":"Logarithms Questions 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