{"id":142840,"date":"2021-08-13T14:04:26","date_gmt":"2021-08-13T08:34:26","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=142840"},"modified":"2021-09-14T15:44:49","modified_gmt":"2021-09-14T10:14:49","slug":"logarithm-questions-for-nmat-2021-examination","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/logarithm-questions-for-nmat-2021-examination\/","title":{"rendered":"Logarithm Questions for NMAT 2021 Examination"},"content":{"rendered":"

Logarithm Questions for NMAT:<\/strong><\/span><\/h1>\n

Download Logarithm Questions for NMAT PDF. Top 10 very important Logarithm Questions for NMAT based on asked questions in previous exam papers.<\/p>\n

Download Logarithm Questions for NMAT<\/a><\/p>\n

Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n

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Question 1:\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 2:\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 3:\u00a0<\/b>If $\\log_{12}{81}=p$, then $3(\\dfrac{4-p}{4+p})$ is equal to<\/p>\n

a)\u00a0$\\log_{4}{16}$<\/p>\n

b)\u00a0$\\log_{6}{16}$<\/p>\n

c)\u00a0$\\log_{2}{8}$<\/p>\n

d)\u00a0$\\log_{6}{8}$<\/p>\n

Question 4:\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

Question 5:\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

<\/a><\/div>\n

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>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 8:\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 9:\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

Question 10:\u00a0<\/b>If x is a positive quantity such that $2^{x}=3^{\\log_{5}{2}}$. then x is equal to<\/p>\n

a)\u00a0$\\log_{5}{8}$<\/p>\n

b)\u00a0$1+\\log_{3}({\\frac{5}{3}})$<\/p>\n

c)\u00a0$\\log_{5}{9}$<\/p>\n

d)\u00a0$1+\\log_{5}({\\frac{3}{5}})$<\/p>\n

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

1)\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

2)\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

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

Given that:\u00a0$\\log_{12}{81}=p$<\/p>\n

$\\Rightarrow$\u00a0$\\log_{81}{12}=\\dfrac{1}{p}$<\/p>\n

$\\Rightarrow$\u00a0$4\\log_{3}{3*4}=\\dfrac{1}{p}$<\/p>\n

$\\Rightarrow$\u00a0$1+\\log_{3}{4}=\\dfrac{4}{p}$<\/p>\n

Using Componendo and Dividendo,<\/p>\n

$\\Rightarrow$ $\\dfrac{1+\\log_{3}{4}-1}{1+\\log_{3}{4}+1}=\\dfrac{4-p}{4+p}$<\/p>\n

$\\Rightarrow$\u00a0$\\dfrac{\\log_{3}{4}}{2+\\log_{3}{4}}=\\dfrac{4-p}{4+p}$<\/p>\n

$\\Rightarrow$\u00a0$\\dfrac{\\log_{3}{4}}{\\log_{3}{9}+\\log_{3}{4}}=\\dfrac{4-p}{4+p}$<\/p>\n

$\\Rightarrow$\u00a0$\\dfrac{\\log_{3}{4}}{\\log_{3}{36}}=\\dfrac{4-p}{4+p}$<\/p>\n

$\\Rightarrow$\u00a0$3*\\dfrac{4-p}{4+p}=\\dfrac{3\\log_{3}{4}}{\\log_{3}{36}}$<\/p>\n

$\\Rightarrow$\u00a0$3*\\dfrac{4-p}{4+p}=\\dfrac{\\log_{3}{64}}{\\log_{3}{36}}$<\/p>\n

$\\Rightarrow$\u00a0$3*\\dfrac{4-p}{4+p}=\\log_{36}{64}$<\/p>\n

$\\Rightarrow$\u00a0$3*\\dfrac{4-p}{4+p}=\\log_{6^2}{8^2}=\\log_{6}{8}$. Hence, option D is the correct answer.<\/p>\n

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

5)\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

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(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

8)\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

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

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

Givne that:\u00a0$2^{x}=3^{\\log_{5}{2}}$<\/p>\n

$\\Rightarrow$ $2^{x}=2^{\\log_{5}{3}}$<\/p>\n

$\\Rightarrow$\u00a0$x=\\log_{5}{3}$<\/p>\n

$\\Rightarrow$\u00a0$x=\\log_{5}{\\dfrac{3*5}{5}}$<\/p>\n

$\\Rightarrow$\u00a0$x=\\log_{5}{5}+\\log_{5}{\\dfrac{3}{5}}$<\/p>\n

$\\Rightarrow$\u00a0$x=1+\\log_{5}{\\dfrac{3}{5}}$. Hence, option D is the correct answer.<\/p>\n

Get 5 NMAT Mocks for Rs. 499<\/a><\/p>\n

<\/a><\/div>\n

We hope this Logarithm Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

Logarithm Questions for NMAT: Download Logarithm Questions for NMAT PDF. Top 10 very important Logarithm Questions for NMAT based on asked questions in previous exam papers. Take NMAT mock test Question 1:\u00a0Sham is trying to solve the expression: $\\log \\tan 1^\\circ + \\log \\tan 2^\\circ + \\log \\tan 3^\\circ\u00a0+ …….. +\u00a0\\log \\tan 89^\\circ$. The correct […]<\/p>\n","protected":false},"author":42,"featured_media":142844,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,4977],"tags":[5015,3131,4979,5013],"class_list":{"0":"post-142840","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-nmat","9":"tag-logarithms-quetions-for-nmat","10":"tag-nmat","11":"tag-nmat-2021","12":"tag-nmat-logarithms"},"better_featured_image":{"id":142844,"alt_text":"NMAT Logarithm Questions","caption":"NMAT Logarithm Questions","description":"NMAT Logarithm 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