MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 7

The value of

$\log_{10} 8$ is equal to

0%
$.903$
0%
$3.901$
0%
$.301$
0%
None of the above

Explanation

$\log_{10}8=\log_{10}2^3=3\log_{10}2$

$=3\times 0.301$

$(\log_{10}2=0.301)$

$=0.903$

Hence, D is the correct option.

The logarithm of number lying between

$0$ and

$1$ is

0%
positive
0%
negative
0%
zero
0%
none of the above

Explanation

From the graph, we can see that logarithmic value of number lying between

$0$ and

$1$ is negative.

Hence, B is the correct option.

Using logarithm table, determine the value of

$\log_{10}0.5432$ .

0%
$\overline {1}.7350$
0%
$\overline {2}.7350$
0%
$0.7350$
0%
$0.07350$

Explanation

$\log_{10}0.5432$

Characteristics

$=-1$

For mantissa ,read from the table

$5432.$

Mantissa

$=7350$

Thus,

$\log_{10}0.5432=$ characteristics of

$0.5432+$ mantissa of

$0.5432$

$=-1+0.7350$

$=\bar1.7350$

Hence, A is the correct option.

Find the value of

$\log_{10} 72$ using log table

0%
$0.901+0.909$
0%
$0.903+0.954$
0%
$1.890$
0%
$2.104$

Explanation

$\log_{10}{72}=\log_{10}(2^{3}.3^{2})$

$=\: \log_{10}{2^{3}}+\log_{10}{3^{2}}$

$=\: 3.\log_{10}{2}+2.\log_{10}{3}$

$=\: 3(0.301)+2(0.4771)$

$=\: 0.903+0.954$

Find the value of

$\log_{10} {\dfrac{64^{2.1}\times 81^{4.2}}{49^{3.4}}}$ using log table

0%
$2.1 \times 6 \times .303+ 4.2 \times 2 \times .854- 3.4 \times 2 \times .745$
0%
$2.1 \times .303+ 4.2 \times .954- 3.4 \times .845$
0%
$2.1 \times 6 \times .303- 4.2 \times 2 \times .954+3.4 \times 2 \times .845$
0%
$2.1 \times 6 \times .303+ 4.2 \times 2 \times .954- 3.4 \times 2 \times .845$

Explanation

$\log_{10}{64^{2.1}}+\log_{10}{81^{4.2}}-\log_{10}{49^{3.4}}$

$= 2.1\log_{10}{64}+4.2\log_{10}{81}-3.4\log_{10}{49}$

$=2.1\log_{10}{2^{6}}+4.2\log_{10}{3^{4}}-3.4\log_{10}{7^{2}}$

$=2.1\times6\times(0.303)+4.2\times4\times(0.477)-3.4\times2\times(0.845)$

$=2.1\times6\times(0.303)+4.2\times2\times(0.954)-3.4\times2\times(0.845)$

Find the value of

${\log_{10} 72} + {\log_{10} {\dfrac{1}{8}}}$ using log table

0%
$0.903$
0%
$0.303$
0%
$0.954$
0%
$1.234$

Explanation

$\log_{10}{72}+\log_{10}\left (\dfrac{1}{8} \right )$

$=\: \log_{10}{\left (72\times \dfrac{1}{8} \right )}=\log_{10}{9}=\log_{10}{3^{2}}$

$=\: 2.\log_{10}{3}=2(0.477)=0.954$

Find the value of

$\dfrac {\log_{10} 72}{\log_{10} 8}$ using log table

0%
$\log_{10} 9$
0%
$1+\dfrac{.954}{.903}$
0%
$2$
0%
$\dfrac{.903+.954}{.954}$

Explanation

$\dfrac{\log_{10}{72}}{\log_{10}{8}}=\dfrac{\log_{10}{(8\times9)}}{\log_{10}{8}}=\dfrac{\log_{10}{8}+\log_{10}{9}}{\log_{10}{8}}$

$=1+\dfrac{\log_{10}{3^{2}}}{\log_{10}{2^{3}}}=1+\dfrac{2\log_{10}{3}}{3\log_{10}{2}}$

$=1+\dfrac{2.(0.477)}{3.(0.301)}=1+\dfrac{(0.954)}{(0.903)}$

$\log 625 = k \log 5$ , then the value of

$k$ is ____

0%
$5$
0%
$4$
0%
$3$
0%
$2$

Explanation

Given,

$\log 625=k \log 5$

$\Rightarrow \log (5^4)=k \log 5$

$\Rightarrow 4\log 5=k \log 5$

$\Rightarrow k=4$

Option

$B$ is correct.

$\dfrac{\log\, x}{l+m-2n} = \dfrac{\log\, y}{m+n-2l} = \dfrac{\log\, x}{n+l-2m}$ , then

$xyz$ is equal to

0%
$0$
0%
$lmn$
0%
$1$
0%
$2$

Explanation

Let

$\dfrac{\log x}{l+m-2n}=\dfrac{\log y}{m+n-2l}=\dfrac{\log z}{n+l-2m}=k(say)$

So, we get

$\log x = k(l+m-2n)$ .......

$(i)$

$\log y = k(m+n-2l)$ .......

$(ii)$

$\log z = k(n+l-2m)$ .......

$(iii)$

$\therefore \log x+\log y + \log z = k(l+m-2n)+k(m+n-2l)+k(n+l-2m)$

$\Rightarrow \log x+\log y + \log z = kl+km-2kn+km+kn-2kl+kn+kl-2km$

$\Rightarrow \log (xyz) = 0$

$\Rightarrow \log xyz = \log 1$

$\Rightarrow xyz = 1$

Simplify the following using law of exponents.

$9^2 \times 9^{18}\times 9^{10}$

0%
$9^{30}$
0%
$9^{25}$
0%
$9^{15}$
0%
$9^{10}$

Explanation

We know that,

$a^{m}\times a^{n}\times a^{p}=a^{m+n+p}$

Therefore,

$9^{2}\times 9^{18}\times 9^{10}=9^{2+18+10}$

$=9^{30}$

Simplify the following using law of exponents for

$a=2, x=1,y=1,z=1$

$a^x\times a^y\times a^z$

0%
$2$
0%
$4$
0%
$8$
0%
$16$

Explanation

we know,

$a^{x}*a^{y}*a^{z}=a^{x+y+z}$

$\therefore2^{1+1+1}$

$=2^{3}$

$=8$

$\log_{x} 484 - \log_{x}4 + \log_{x}14641 - \log_{x}1331 = 3$ , then the value of

$x$ is

0%
$1$
0%
$3$
0%
$11$
0%
None of these

Explanation

$\log_{x} 484 - \log_{x}4 + \log_{x}14641 - \log_{x}1331 = 3$

$\Rightarrow \log_{x}\left (\dfrac {484}{4}\right ) + \log_{x} 14641 - \log_{x} 1331 = 3$

$\Rightarrow \log_{x} 11^{2} + \log_{x} 11^{4} - \log_{x} 11^{3} = 3$

$\Rightarrow (2 + 4 - 3)\log_{x} 11 = 3\Rightarrow \log_{x}11 = 1$

$\Rightarrow 11$

Hence choice (c) is correct.

$\log_8 {64}$ is equal to_____

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\log_8(64)$

$=\log_8(8^2)$

$=2\log_88$

$=2$ (

$\because \log_aa=1)$

Hence, A is the correct option.

$\log_4{64}$ is equal to____

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$\log_4(64)$

$=\log_4(4^3)$

$=3\log_44$

$=3$ (

$\because \log_aa=1)$

Hence, B is the correct option.

The value of

$[2-3(2-3)^{-1}]^{-1}$ is __________.

0%
$\displaystyle\frac{-1}{5}$
0%
$\displaystyle\frac{1}{5}$
0%
$-5$
0%
$5$

Explanation

$=$

$[2-3(2-3)^{-1}]^{-1}$

$=$

$[2-3(-1)^{-1}]^{-1}$

$=$

$[2-3(-1)]^{-1}$

$=$

$[2+3]^{-1}$

$=$

$\dfrac{1}{5}$

Which of the following values are equal?
(P)

$1^{4}$ (Q)

$4^{\circ}$ (R)

$0^{4}$ (S)

$4^{1}$ .

0%
$P$ and $Q$
0%
$Q$ and $R$
0%
$P$ and $R$
0%
$P$ and $S$

Explanation

(P)

$1^{4} = 1$

(Q)

$4^{0} = 1$

(R)

$0^{4} = 0$

(S)

$4^{1} = 4$

Hence,

$P$ and

$Q$ are equal.

$\log_3 {27}$ is equal to____

0%
$3$
0%
$2$
0%
$4$
0%
$5$

Explanation

$\textbf{Step 1 : Using the principle properties of logarithm}$

$\boldsymbol{\log_a m^n = n \log_a m\, ; \, where \,\,m, a > 0,\neq1}$

$\text{Given : }\log_3 27$

$= \log_3 3^3$

$\textbf{Step 2 : Using the principle properties of logarithm}$

$\boldsymbol{\log_a a = 1 ; \, where \,\,a > 0,\neq1}$

$= 3 \log_3 3 = 3$

$\log {5} + \log {8}$ is equal to........

0%
$\log {40}$
0%
$\log{13}$
0%
$\dfrac{\log{8}}{\log{5}}$
0%
$\log{3}$

Explanation

$\textbf{Step 1 : Use properties of logarithm }$

$\text{Given , }\log 5+\log8$

$=\log(5\times8)$

$\boldsymbol{[\because \log a+\log b =\log(a \times b)]}$

$=\log 40$

$\textbf{Hence, A is the correct option.}$

$\log {3} + \log {5}$ is equal to..........

0%
$\log {15}$
0%
$\log {8}$
0%
$\dfrac{\log {3}}{\log{5}}$
0%
$\log{2}$

Explanation

$\textbf{Step 1 : Use the properties of log }$

$\text{Given , }\log 3+\log5$

$=\log(3\times5)$ [

$\boldsymbol{\because \log a+\log b =\log(a \times b)}$ ]

$=\log 15$

$\textbf{Hence, A is the correct option.}$

$x^2+y^2=25$ , then

$log_5 \begin {bmatrix} Max (3x+4y) \end {bmatrix}$ is

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$log_5(3x+4y)$

Let

$s=3x+4y$

given

$x^2+y^2=25$

then

$S=3x+4 \sqrt{25-x^2}$

$\dfrac{ds}{dx}=3+4 \dfrac{1}{2\sqrt(25-x^2)}$

$3=\dfrac{4x}{\sqrt{25-x^2}}$

$9(25-x^2)=16x^2$

$x=\pm 3$

and

$x^2+y^2=25$

$y^2=25-x^2$

$y=\pm 4$

$\dfrac{d^2s}{dx^2}<0$ ; At

$x=3 \,and\, y=4$

$S=3x+4y=3(3)+4(4)=25$

$log_5 (3x+4y)=log_5(s)=log_5(25)=log_5(5^2)=2$

$\log_2 {64}$ is equal to____

0%
$2$
0%
$3$
0%
$4$
0%
$6$

Explanation

$\log_2(64)$

$=\log_2(2^6)$

$=6\log_22$

$=6$ (

$\because\log_aa=1)$

Hence, D is the correct option.

Simplify:

$(-1)^{19}+(-1)^{20}+(2)^5$

0%
$34$
0%
$32$
0%
$30$
0%
$0$

Explanation

$(-1)^{19}+(-1)^{20}+(2)^5$

$=(-1)+(1)+2\times2\times2\times2\times2$

$=32$

$\displaystyle \int{ \frac{(\sqrt{x})^5}{(\sqrt{x})^7 + x^6} } dx = a\ log \left( \frac{x^k}{1 + x^k} \right) + c$ , then a and k are

0%
$\displaystyle \frac{2}{5}, \frac{5}{2}$
0%
$\displaystyle \frac{1}{5}, \frac{2}{5}$
0%
$\displaystyle \frac{5}{2}, \frac{1}{2}$
0%
$\displaystyle \frac{2}{5}, \frac{1}{2}$

Explanation

Given

$\int \dfrac{(\sqrt{x})^5}{(\sqrt{x})^7+x^6} dx = a log \left ( \dfrac{x^k}{1+x^k} \right )+c$

L.H.S Taking

$x^6$ common from denominator

$\displaystyle \Rightarrow \int \dfrac{(\sqrt{x})^5}{x^6 \left ( \dfrac{(\sqrt{x})^7}{x^6}+1 \right )}dx$

$\displaystyle \Rightarrow \int \dfrac{1}{(\sqrt{x})^7 (\dfrac{1}{\sqrt{x}})^5 + 1 }dx$

Now Let

$\dfrac{1}{(\sqrt{x})^5} = t$

$\dfrac{-5}{2(\sqrt{x})^7} .dx = dt$

$\displaystyle \Rightarrow \int \frac{-2}{3}.\frac{1}{(t+1)}dt$

$\Rightarrow \dfrac{-2}{5} ln (t+1)+c$

put

$t = \dfrac{1}{(\sqrt{x})^5}$

$\Rightarrow \dfrac{-2}{5} ln \left ( \dfrac{1+(\sqrt{x})^5}{(\sqrt{x})^5} \right )+c \Rightarrow \dfrac{2}{5} ln \left ( \dfrac{(\sqrt{x})^5}{1+(\sqrt{x})^5} \right ) +c$

on comparing

$\boxed{a = t\dfrac{2}{5}} \boxed {k = 5/2}$

1154012_876314_ans_03a7fca2e12345c5b5d697938ea561c2.jpg

If the mantissa of

$\log 2125 =3.3273$ , find the mantissa of

$\log21.25$

0%
$1.3273$
0%
$2.3273$
0%
$0.3273$
0%
$32.2321$

The logarithm of

$0.0625$ to the base

$2$ is:

0%
$0.025$
0%
$0.25$
0%
$5$
0%
$-4$
0%
$-2$

Explanation

$\log_{2}{\cfrac{625}{10000}}$

$=\log_{2}{\cfrac{1}{16}}$

$=\log_{2}{(16)^{-1}}$

$=\log_{2}{(2)^{-4}}$

$=-4\log_{2}{2}$

$= -4$

The equation

$x^{\log_{\sqrt{x}}{2x}} = 4$ has no solution.

0%
True
0%
False

Explanation

$\textbf{Step 1: Exponential properties based on base change theorem:}$

$\boldsymbol{a^{\log_bc} = c^{\log_ba}}$

$\textbf{(Where a,b,c > 0}$ &

$\textbf{b} \neq 1)$

$\textbf{Given:}$

$x^{\log_{\sqrt{x}}{2x}} = 4$

$\Rightarrow (2x)^{\log_{\sqrt{x}}{x}} = 4$ ...............(

$\because \boldsymbol{a^{\log_bc} = c^{\log_ba}}$ )

$\textbf{Step 2: The principle properties of logarithm:}$

$\boldsymbol{ \log_{a^p}m = \dfrac{1}{p} \log_am}$ (

$\textbf{Where a,m > 0 & a} \neq 1)$

$\Rightarrow (2x)^{\dfrac{1}{1/2} \log_xx} = 4$ ...............(

$\because \boldsymbol{ \log_{a^p}m = \dfrac{1}{p} \log_am}$ )

$\Rightarrow (2x)^2 = 4$

$\Rightarrow 4x^2 = 4 \Rightarrow x^2 = 1$

$\Rightarrow x = \pm 1$ (

$x = -1 , 1$ are rejected as log base cant be 1 and number must be positive )

$\Rightarrow No Solution$

$t = -2$ then

$log_4 \dfrac{t^2}{4}-2 \, log_4 \, 4t^4 \, =$

0%
2
0%
-4
0%
-6
0%
0

Explanation

$\textbf{Step 1 : The principle properties of logarithm:} \log_am^n = n\log_am \left( \textbf{Where m,a>0 and a}\neq 1\right)$

Consider,

${{\log }_{4}}\dfrac{{{t}^{2}}}{4}-2{{\log }_{4}}4{{t}^{4}}$

Put $t=-2$ in above expression, We get

$={{\log }_{4}}\dfrac{{{\left( -2 \right)}^{2}}}{4}-2{{\log }_{4}}4{{\left( -2 \right)}^{4}}$

$={{\log }_{4}}\left( \dfrac{4}{4}\right)-2{{\log }_{4}}\left(4\times 16\right)$

$={{\log }_{4}}1-2{{\log }_{4}}64$

$=0-2{{\log }_{4}}{{4}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}1=0 \right)$

$=-2\times 3{{\log }_{4}}4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\log }_{a}}{{m}^{n}}=n{{\log }_{a}}m \right)$

$=-6\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \therefore {{\log }_{a}}a=1 \right)$

Hence, $-6$ is the answer.

If mantissa of logarithm of 719.3 to the base 10 is 0.8569 , then mantissa of logarithm of 71.93 is

0%
0.8569
0%
$\overline 1 .8569$
0%
1.8569
0%
0.1431

Explanation

Mantissa of logarithm of

$719.3$ to base

$10$ is

$0.8569$

Then, mantissa of logarithm of

$71.93$ is also

$0.8569$

As,

$\log_{10}{(719.3)}=2+(0.8569)$ (mantissa)

So,

$\log_{10}{(71.93)}=1+(0.8596)$ (mantissa)

$2y = log(12-5x-3x^2)$ takes all real values then

$x$ belongs to

0%
$(-3, 5/3)$
0%
$(-3, 3)$
0%
$(-3, 4/3)$
0%
None

Explanation

$2y=\log\left(12-5x-3x^2\right)$

$(12-5x-3x^2)>0$

$3x^2+5x-12<0$

$3x^2+9x-4x-12<0$

$3x\left(x+3\right)-4\left(x+3\right)<0$

$\left(x+3\right)\left(3x-4\right)<0$

$x\epsilon \left(-3,{4/3}\right)$

The value of

${\log _b}a.{\log _c}b.{\log _a}c$ is

0%
0
0%
1
0%
${\log _a}abc$
0%
10

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 7 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 7 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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