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# 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.

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Question 1:Â Sham is trying to solve the expression:
$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circÂ + …….. +Â \log \tan 89^\circ$.

a)Â 1

b)Â $\frac{1}{\sqrt{2}}$

c)Â 0

d)Â -1

Question 2:Â Find the value of $\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

a)Â $n^{2} + 1$

b)Â $n^{2} – 1$

c)Â $\frac{(n^{2} + n)}{2}.\frac{n(n + 1)}{3}$

d)Â $\frac{(n^{2} + n)}{2}$

Question 3:Â If $\log_{12}{81}=p$, then $3(\dfrac{4-p}{4+p})$ is equal to

a)Â $\log_{4}{16}$

b)Â $\log_{6}{16}$

c)Â $\log_{2}{8}$

d)Â $\log_{6}{8}$

Question 4:Â If $\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-B$ and $\log_2{a}=\frac{1}{3}$, then $\log_3{a}$ equals

a)Â $\frac{2}{A+B-3}$

b)Â $\frac{2}{A+B}-3$

c)Â $\frac{A+B}{2}-3$

d)Â $\frac{A+B-3}{2}$

Question 5:Â $(1+5)\log_{e}3+\frac{(1+5^{2})}{2!}(\log_{e}3)^{2}+\frac{(1+5^{3})}{3!}(\log_{e}3)^{3}+…$

a)Â 12

b)Â 244

c)Â 243

d)Â 245

Question 6:Â If $\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$. Then x is

a)Â 2

b)Â 4

c)Â 16

d)Â 12

Question 7:Â Let $u = ({\log_2 x})^2 – 6 {\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has

a)Â no solution for x

b)Â exactly one solution for x

c)Â exactly two distinct solutions for x

d)Â exactly three distinct solutions for x

Question 8:Â If $\log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression, then x is equal to

a)Â $\frac{8}{3}$

b)Â $\log_{3}{8}$

c)Â $\log_{2}{3}$

d)Â $8$

Question 9:Â If Y is a negative number such that $2^{Y^2({\log_{3}{5})}}=5^{\log_{2}{3}}$, then Y equals to:

a)Â $\log_{2}(\frac{1}{5})$

b)Â $\log_{2}(\frac{1}{3})$

c)Â $-\log_{2}(\frac{1}{5})$

d)Â $-\log_{2}(\frac{1}{3})$

Question 10:Â If x is a positive quantity such that $2^{x}=3^{\log_{5}{2}}$. then x is equal to

a)Â $\log_{5}{8}$

b)Â $1+\log_{3}({\frac{5}{3}})$

c)Â $\log_{5}{9}$

d)Â $1+\log_{5}({\frac{3}{5}})$

$\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$.

=$\log \tan 1^\circ + \log \tan 89^\circ + \log \tan 2^\circ + \log \tan 88^\circ …….. + \log \tan 45^\circ$.

=$\log\ \left(\tan\ 1^0\cdot\tan\ 89^0\right)\times\log\ \left(\tan\ 2^0\cdot\tan\ 88^0\right)\ ………………………\log\ \left(\tan\ 45^0\right)$

tan $45^0$ = 1

$\log\ \left(\tan\ 45^0\right)\ =\ 0$

$\therefore$Â $\log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + …….. + \log \tan 89^\circ$ = 0

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$

SinceÂ $\log_aa\$ = 1

$\log_{10}{10} + \log_{10}{10^2} + ….. + \log_{10}{10^n}$ = 1+2+….n

=$\ \frac{\ n\left(n+1\right)}{2}$

=$\frac{(n^{2} + n)}{2}$

Given that:Â $\log_{12}{81}=p$

$\Rightarrow$Â $\log_{81}{12}=\dfrac{1}{p}$

$\Rightarrow$Â $4\log_{3}{3*4}=\dfrac{1}{p}$

$\Rightarrow$Â $1+\log_{3}{4}=\dfrac{4}{p}$

Using Componendo and Dividendo,

$\Rightarrow$ $\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$

$\Rightarrow$Â $\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$Â $\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$Â $\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$

$\Rightarrow$Â $3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$

$\Rightarrow$Â $3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$

$\Rightarrow$Â $3*\dfrac{4-p}{4+p}=\log_{36}{64}$

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

$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$………..(1)

$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$………….(2)

and finally $\log_a2=3$

Substituting this in (1) we get $\log_a5+\log_a3=A-3$

Now we have two equations in two variables (1) and (2) . On solving we get

$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$

Splitting the above mentioned series into two series

A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$

B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$

We know that $e^{x}$ =$1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$

SoÂ Â $e^{x}-1$ = $x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$

On solving two series A and B

A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$ =$e^{\log_{e}3}-1$ = $3-1$ =$2$

B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$=$e^{\log_{e}3^{5}}-1$=$3^{5}-1$=$242$

A+B = $2 + 242$ = $244$

$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$

i.e. $\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$

i.e.Â $\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\log{2}$

let t = log x

Therefore,Â Â $\frac{t * (t-log{16})}{t-log{64}}$ = $\log{2}$

$t^2-4*log 2*t = t*log 2-6*(log 2)^2$

I.e.Â $t^2-5*log 2*t-6*(log 2)^2$ = 0

I.e.Â $t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0

i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0

i.e $t=2*log 2$ or $t=3*log 2$

i.e $log x=log 4$ or $log x=log 8$

therefore $x=4$ or $8$

therefore our answer is option ‘B’

$x^u = 256$

Taking log to the base 2 on both the sides,

$u * \log_{2}{x} = \log_{2}{256}$

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

$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$

Let $log_2 x = t$

$t^3 – 6t^2 +12t – 8 = 0$

$(t-2)^3 = 0$

Therefore, $log_2 x = 2$

=> $x = 4$ is the only solution

Hence, option B is the correct answer.

If $log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression
Then, $2*log(3^{x} – 2) = log{3}+log (3^{x}+ 4)$
Thus, $log{(3^{x} – 2)^2} = log{3(3^x+4)}$
Thus, $(3^{x} – 2)^2 = 3(3^x+4)$
=> $3^{2x} – 4*3^x +4 = 3*3^x + 12$
=> $3^{2x} – 7*3^x – 8 = 0$
=> $(3^x+1)*(3^x-8) = 0$
But $3^x+1 \neq 0$
Thus, $3^x = 8$
Hence, $x = log_{3}{8}$
Hence, option B is the correct answer.

$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$

Given,Â $5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$

=>Â $Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$

=>$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$

since Y is a negative number, Y=$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$

Givne that:Â $2^{x}=3^{\log_{5}{2}}$

$\Rightarrow$ $2^{x}=2^{\log_{5}{3}}$

$\Rightarrow$Â $x=\log_{5}{3}$

$\Rightarrow$Â $x=\log_{5}{\dfrac{3*5}{5}}$

$\Rightarrow$Â $x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$

$\Rightarrow$Â $x=1+\log_{5}{\dfrac{3}{5}}$. Hence, option D is the correct answer.

We hope this Logarithm Questions for NMAT pdf for NMAT exam will be highly useful for your Preparation.