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Logarithm and its uses - class-XI

Description: logarithm and its uses
Number of Questions: 78
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Tags: basic mathematical concepts physics
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$\log _ee^5$ is equal to- 

  1. $2.5$

  2. $1.5$

  3. $2$

  4. $5$


Correct Option: D
Explanation:

$\log _ee^5=5\log _ee=5*1=5$

If $\log _3 x = 3\, & \,\log _x y = 4\,$, then find $y$.

  1. ${3^6}$

  2. ${3^9}$

  3. ${3}^{12}$

  4. $none$


Correct Option: C
Explanation:

We have,

${{\log } _{3}}x=3$

 

We know that

${{\log } _{a}}x=y\Rightarrow x={{a}^{y}}$

 

Therefore,

$ x={{3}^{3}} $

$ x=27 $

 

Here,

$ {{\log } _{x}}y=12 $

$ y={{x}^{12}} $

 

On putting the value of $x$, we get

$y={{3}^{12}}$

 

Hence, this is the answer.

If anti ${ \log } _{ 10 }(0.3678)=2.3324$ then ${ \log } _{ 10 }233.2$ is equal to

  1. 367.8

  2. 36.78

  3. 3.3678

  4. 2.3678


Correct Option: D
Explanation:

$antilog _{10}(0.3678)=2.3324$


$\therefore \log(2.3324)=0.3678$

$\log _{10}233.2=\log _{10}(2.332\times 100)$

                     $=\log _{10}(2.332) + \log (10^2) \quad \dots ( \log a+\log b=\log(ab))$

                     $=\log _{10}(2.332) + 2\log (10) \quad \dots ( n \log _ab=\log _ab^n)$

As we know, $log _a a=1$

$\log _{10}233.2=2+0.3678=2.3678$

$\log _264+\log _3729$

  1. 12

  2. 15

  3. 18

  4. None


Correct Option: A
Explanation:

$\log _264+\log _3729\\log _22^6+\log _33^6\6\log _22+6\log _33\6+6=12$

$\log _5625+\log _6 216$

  1. $7$

  2. $6$

  3. $5 $

  4. $9$


Correct Option: A
Explanation:

$\log _5625+\lg _6216\\log _55^4+\log _66^3\4\log _55+3\log _66\3+4=7$

Which of the following real numbers is(are) non-positive?

  1. $log{ } _{ 0.3 }(\dfrac { \sqrt { 5 } +2 }{ \sqrt { 5 } -2 }  )$

  2. $log{ } _{ 7 }(\sqrt { 83 } -9\quad )$

  3. $log{ } _{ 7\frac { \pi }{ 12 } }(cot\frac { \pi }{ 8 } \quad )$

  4. ${ log } _{ 2 }\sqrt { 9.\sqrt [ 3 ]{ { 27 }^{ \frac { -5 }{ 3 } }.243{ }^{ \frac { -7 }{ 5 } } } } $


Correct Option: A
Explanation:
From given,

we have,

$\log _{0.3}\left(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\right)$

$=\log _{0.3}\left(2+\sqrt{5}\right)-\log _{0.3}\left(\sqrt{5}-2\right)$

$=-2.39811$

If $x=\log _am$, then value of $m$ is equal to

  1. anti log $ _a x$

  2. anti log $ _x a$

  3. $a^x$

  4. $x^a$


Correct Option: A,C
Explanation:

We have, 

$x = \log _am $
$\Rightarrow a^x = m $
or $\therefore m = $ $\text{anti}\log _ax$

If $\log x = -2.0258$, then $x$ is equal to

  1. $0.009223$

  2. $0.009423$

  3. $0.008422$

  4. $0.008223$


Correct Option: B
Explanation:

$\log x=-2.0258$

Characteristics$=-3$
Mantissa$=-2.0258-(-3)=0.9742$
Value of $0.9742$ from antilog table $=9419+4=9423$
Number of zeroes placed after decimal will be $2.$
Antilog $-2.0258=0.009423$
Hence, B is the correct option.

The antilog of the number $2.5463$ is

  1. $251.8$

  2. $254.8$

  3. $351.8$

  4. $354.8$


Correct Option: C
Explanation:

The number before decimal point is $2,$ so decimal point will be after $3$ digits.

Value of $0.5463$ from antilog table $=3516+2=3518$
Now place a decimal point after $3$ digits of the number from left we get, $351.8$
Antilog $2.5463=351.8$
Hence, C is the correct option.

The antilog of $ (.2817)$ will be

  1. $1.19$

  2. $0.91$

  3. $0.19$

  4. $1.91$


Correct Option: D
Explanation:

The number before decimal point is $0,$ so decimal point will be after $1$ digits.

Value of $0.2817$ from antilog table $=1910+3=1913$
Now place a decimal point after $1$ digit of the number from left we get, $1.913$
Antilog $0.2817=1.913=1.91$
Hence, D is the correct option.

What is the value of $[\log _{10} (5\log _{10} 100)]^{2}$?

  1. $4$

  2. $3$

  3. $2$

  4. $1$


Correct Option: D
Explanation:

The value of $[\log _{10}(5\log _{10} 100)]^{2}$ is
$=[\log _{10}(5\log _{10}10^2)]^2$

$=[\log _{10}(10\log _{10}10)]^2$     .....As $\log a^m=m\log a$
$=[\log _{10}(10\times 1)]^2$     ....As $\log _aa=1$
$=[\log _{10}10]^2$
$=[1]^2=1$

Find the characteristic of $\log 7.93$

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: A
Explanation:

From Logarithmic table,

$\log7.93=0.89927$
Here, Characteristics$=0$
Hence, A is the correct option.

Find the characteristic of $\log 277.9301$

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: C
Explanation:

From logarithmic table,

$\log 277.9301=2.4439$
Here, Characteristics$=2$
Hence, C is the correct option.

Find the characteristic of $\log 27.93$

  1. $0$

  2. $1$

  3. $2$

  4. $3$


Correct Option: B
Explanation:

From log table,

$\log27.93=1.4460$
Here, Characteristics$=1$ and Mantisa$=0.4460$
Hence, B is the correct option.

Find the mantissa of $\log 2.125$

  1. $1.3273$

  2. $2.3273$

  3. $0.3273$

  4. $32.2321$


Correct Option: C
Explanation:

From Logarithmic table,

$\log2.125=0.3237$
Here, Characteristics$=0,$ Mantissa$=0.3237$
Hence, C is the correct option.

The value of $x$ which satisfy $log(x+1) = 2logx$ is 

  1. $1$

  2. $\dfrac{\sqrt{5}-1}{2}$

  3. $\dfrac{\sqrt{5}+1}{2}$

  4. $2$


Correct Option: B,C
Explanation:

$\log(x+1)=2\log x$


$\Rightarrow \log (x+1)=\log x^2$

$\Rightarrow x^2-x-1=0$

$\Rightarrow x=\cfrac{+1\pm \sqrt{1+4}}{2}=\cfrac{1\pm \sqrt{5}}{2}$

$\Rightarrow x= \left(\cfrac{1+\sqrt{5}}{2}\right),\left(\cfrac{1-\sqrt{5}}{2}\right)$

The value of $x$ satisfying the equation $g^{log _3 (log _2 x)} = log _2 x - (log _2 x)^2 + 1$ is 

  1. $0$

  2. $1$

  3. $2$

  4. None


Correct Option: C
Explanation:

Here, $g^{\log _{3}(\log _{2}x)}$ is an exponential function and $\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1$ is a quadratic with imaginary roots.

The two can be equal when both side become $0,1$. Since, right hand side can become zero at imaginary point. We, only consider, then the two side become $1$.
$\log _{2}{x}-\left(\log _{2}{x}\right)^{2}+1=1$
$\Rightarrow \left(\log _{2}{x}\right)^{2}-\left(\log _{2}{x}\right)=0$
$\Rightarrow \left(\log _{2}{x}\right)\left(\log _{2}{x}-1\right)=0$
$\Rightarrow \log _{2}{x}=0$ and $\log _{2}{x}=1$
$\Rightarrow x=1,2$
But $x\neq 1$ as in $g^{\log _{3}(\log _{2}x)}$ it become invalid hence, $x=2$ satisfy the relation.

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

  1. $(-3, 5/3)$

  2. $(-3, 3)$

  3. $(-3, 4/3)$

  4. None


Correct Option: C
Explanation:

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

$(12-5x-3x^2)>0$
$3x^2+5x-12<0$
$3x^2+9x-4x-12<0$
$3x\left(x+3\right)-4\left(x+3\right)<0$
$\left(x+3\right)\left(3x-4\right)<0$
$x\epsilon \left(-3,{4/3}\right)$

Evaluate the expression by using logarithm tables: $ \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$

  1. $11.01$

  2. $12.01$

  3. $13.01$

  4. $14.01$


Correct Option: A
Explanation:

Let $x= \dfrac{(17.42)^{2/{3}}\times 18.42}{\sqrt{126.37}}$
Taking logarithm on both sides,
$ \log { x } =\log { (17.42)^{ 2/{ 3 } } } +\log { 18.42 } -\log { \sqrt { 126.37 }  } $

$\log { x } =\dfrac { 2 }{ 3 } \log { 17.42 } +\log { 18.42 } -\dfrac { 1 }{ 2 } \log { (126.37) } $

$ \log { x } =\dfrac { 2 }{ 3 } \log { (1.742\times 10) } +\log { (1.842\times 10) } -\dfrac { 1 }{ 2 } \log { (1.264\times { 10 }^{ 2 }) } $

$ \log { x } =\dfrac { 2 }{ 3 } { (1.2410) } + { 1.2653 } -\dfrac { 1 }{ 2 } { (2.1018) } $

$\log { x } =0.8273+1.2653-1.0509$

$\log { x } =1.0417$
$\Rightarrow x= \text{antilog }(1.0417)$
$\Rightarrow x = 11.01$

Let $a = \log 3\log _32$. An integer k satisfying  $1< 2^{(-k+3^{-a})} < 2,$  must be less than ____.

  1. $1.25766$

  2. $2.256$

  3. $3$

  4. $1$


Correct Option: A

If $a=\log _35 $ and $b= \log _725$ then correct option is:

  1. $a < b$

  2. $ a > b$

  3. $a= b$

  4. None of these


Correct Option: A
Explanation:

$a=\log _{ 3 }{ 5 } =\cfrac { \log { 5 }  }{ \log { 3 }  } ,b=\cfrac { \log { 25 }  }{ \log { 7 }  } =\cfrac { \log { 5^2 }  }{ \log { 7 }  }=\cfrac { 2\log { 5 }  }{ \log { 7 }  } $


$ \cfrac { a }{ b } =\cfrac { \log { 5 }  }{ \log { 3 }  } \times \cfrac { \log { 7 }  }{ 2\log { 5 }  } =\cfrac { 1 }{ 2 } \log _{ 3 }{ 7 } =\log _{ 3 }{ (\sqrt { 7 } ) } $


$ Now,\sqrt { 7 } <3,so\quad \cfrac { a }{ b } <1$

$ \cfrac { a }{ b } <1\  =>a<b$

The value of ${ \left( 0.05 \right)  }^{ \log _{ \sqrt { 20 }  }{ \left( 0.1+0.01+0.001+.... \right)  }  }$ is 

  1. $81$

  2. $\cfrac{1}{81}$

  3. $20$

  4. $\cfrac{1}{20}$


Correct Option: A
Explanation:
$0.1+0.01+0.0001+..... \Rightarrow G.P$
$=(0.1)(1-0.1)\quad S _{\infty}=-\dfrac {a}{1-r}$
$=\dfrac {0.1}{0.9}=\dfrac {1}{9}$
$\therefore \ (0.05)\log _\sqrt {20} (0.1+0.01+....)$
$=\left (\dfrac {1}{20}\right)\log \sqrt {20}^{1/9}$
$=\left (\dfrac {1}{9}\right) \log \sqrt {20}^{1/20}$
$=\left (\dfrac {1}{20}\right) \log \sqrt {20}^{(\sqrt {20})^{-2}}=\left (\dfrac {1}{20}\right)^{-2}$
$=81$

The equation ${ x }^{ \cfrac { 3 }{ 4 } { \left( \log _{ x }{ x }  \right)  }^{ 2 }+\log _{ x }{ x } -\cfrac { 5 }{ 4 }  }=\sqrt { 2 } $ has

  1. at least one real solution

  2. exactly three solutions

  3. exactly one irrational solution

  4. complex roots


Correct Option: A

If $\log _{10}e=0.4343$, then $\log _{10}1016$ is

  1. $2.99$

  2. $3$

  3. $3.006949$

  4. $3.02$


Correct Option: C
Explanation:

$\log _{10}1016\Rightarrow \dfrac{\log 1016}{\log 10}$

$\Rightarrow \dfrac{3.006949}{1}$
$\rightarrow$ Option $C$ is correct

Multiple Correct:

Which of the following statements are true

  1. $\log _{ 2 }{ 3 } <\log _{ 12 }{ 10 } $

  2. $\log _{ 6 }{ 5 } <\log _{ 7 }{ 8 } $

  3. $\log _{ 3 }{ 26 } <\log _{ 2 }{ 9 } $

  4. $\log _{ 16 }{ 15 } >\log _{ 10 }{ 11 } >\log _{ 7 }{ 6 }$


Correct Option: B,C

The solution of the equation $\log _{7}\log _{5}(\sqrt {x^{2}}+5+x)=0$

  1. $x=2$

  2. $x=3$

  3. $x=0$

  4. $x=-2$


Correct Option: C
Explanation:

$\log _{7}\log _{5}(\sqrt{x^{2}}+5+x)=0$

$\log _{5}(\sqrt{x^{2}}+5+x)=1$
$\sqrt{x^{2}+5+x=5}$
$\sqrt{x^{2}+x=0}$
$x^{2}+x^{2}+2\sqrt[x]{x^{2}}=0$
$x=0$


The value of $\displaystyle\sum _{r=1}^{n}log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$ is

  1. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n}} \right )$

  2. $\dfrac{n}{2}log\left ( \dfrac{a^{n}}{b^{n+1}} \right )$

  3. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n+1}} \right )$

  4. $\dfrac{n}{2}log\left ( \dfrac{a^{n+1}}{b^{n-1}} \right )$


Correct Option: D
Explanation:
Now,
$\displaystyle\sum _{r=1}^{n}\log\left ( \dfrac{a^{r}}{b^{r-1}} \right )$
$=\displaystyle\sum _{r=1}^{n}\left(\log a^{r}-\log b^{r-1}\right)$
$=\displaystyle\sum _{r=1}^{n}\left(r\log a-(r-1)\log b\right)$
$=(\log a)\times \dfrac{n(n+1)}{2}-\log b\times\dfrac{(n-1)n}{2}$
$=\dfrac{n}{2}\left(\log a^{n+1}-\log b^{n-1}\right)$
$=\dfrac{n}{2}\log\left(\dfrac{a^{n+1}}{b^{n-1}}\right)$

Find the value of $\log _{10}{\left(0.\bar{9}\right)}$

  1. $0$

  2. $1$

  3. $-1$

  4. $2$


Correct Option: A
Explanation:
To find value of $\log _{10}{0.\bar9}$

Let $x=0.\bar{9}=0.999999...$

$\Rightarrow 10x=9.99999....$

$\Rightarrow 10x-x=9$

$\Rightarrow 9x=9$

$\Rightarrow x=\dfrac{9}{9}=1$

$\therefore x=1$

Let $y=\log _{10}{1}$

$\Rightarrow 1={10}^{y}$

$\Rightarrow {10}^{y}={10}^{0}$                 (since ${10}^{0}=1$)

Since bases are same we can equate the powers

$\therefore y=0$

Hence, $\log _{10}{\left(0.\bar{9}\right)}=0$

Given $log2=a,log3=b$ express the following in terms of $a$ or $b$ or both

  1. $\log1.5$

  2. $ \log1.2$

  3. $\log0.24$

  4. $ \log0.5$

  5. $\log0.036$


Correct Option: A

If $y=a\log\left|x\right|+bx^{2}+x$ has extreme values at $x=2$ and $x=-4/3$ then 

  1. $a=12,b=-10$

  2. $a=4,b=-3/4$

  3. $a=-6,b=1/4$

  4. $none$


Correct Option: A

Let $A=\dfrac{1}{6}((\log _{2}{3}))^{3}-(\log _{2}{6}))^{3}-(\log _{2}{12}))^{3}+(\log _{2}{24}))^{3})$. Then the value of $2^{A}$ is :

  1. $72$

  2. $70$

  3. $68$

  4. $None\ of \ these$


Correct Option: A

If $x=500,y=100$ and $z=5050$, then the value of $(\log _{ xyz }{ { x }^{ z } } )(1+\log _{ x }{ yz } )$ is equal to.

  1. 500

  2. 100

  3. 5050

  4. 10


Correct Option: C
Explanation:
Given,

$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$

$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$

$=z\log _{xzy}\left(x\right)\left(\log _x\left(zy\right)+1\right)$

from given w have,

$=5050\log _{(500 \times 5050 \times 100)}\left(500\right)\left(\log _{500}\left((5050 \times 100)\right)+1\right)$

$=\dfrac{5050\log _e \left(505000\right)}{\log _e \left(252500000\right)}+5050\log _{252500000}\left(500\right)$

$=5050$

The value of $(0.2)^{log _{\sqrt{5}} \left(\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + ...\right)}$ is

  1. $1$

  2. $2$

  3. $\dfarc{1}{2}$

  4. $4$


Correct Option: A

Find the mantissa of the logarithm of the number $0.002359$.

  1. $3710$

  2. $3718$

  3. $3728$

  4. $3742$


Correct Option: C
Explanation:

We have to find $\log (0.002359)$

Firstly, we will write $0.002359$ in standard form.
So, $0.002359 = 2.359 \times 10^{-3}$
Here, characteristic is -3.

To find the mantissa of $\log (0.002359)$, we first look in the row starting with 23. In this row, look at the number in the column headed by 5. The number is 3711.

Now, move to the column of mean differences and look under the column headed by 9 in the row corresponding to 23. We see the number 17 here.

Add this number to 3711. We get the number 3728. This is the required mantissa of $\log (0.002359)$.

Mantissa of $\log 23.598$, $\log 2.3598$ and 0.023598 is the same (only characteristics are different).

The domain of the function $f(x)=[log _{10}(\frac{5x-x^2}{4})]^{{1}/{2}}$ is 

  1. $- \infty < x < \infty $

  2. $1\le x \le 4$

  3. $4\le x \le 16$

  4. $-1\le x \le 1$


Correct Option: B
Explanation:
$log _{10}\dfrac{5x-x^2}{4}>0$
$\dfrac{5x-x^2}{4}\geq 1$
$5x-x^2\geq 4$
$x^2-5x+4 \leq 0$
$(x-4)(x-1) \leq 0$
$x $ belongs to $[1,4]$
So the domain is $1 \leq x \leq 4$

If $A=log _2 log _2 log _4 256+2 log \sqrt { 2 } 2$ then A=

  1. $2$

  2. $3$

  3. $5$

  4. $7$


Correct Option: C

The value of $\displaystyle \log _{\frac{1}{20}}40$ is

  1. greater than zero.

  2. smaller than zero.

  3. greater than zero and smaller than one.

  4. none of these


Correct Option: B
Explanation:

Let,  $y=\displaystyle \log _{\frac{1}{20}}40$
$\Rightarrow y=-\displaystyle \log _{20}40 \quad [\because \log _{1/a}b=-\log _ab]$
Now $20 < 40<20^2\Rightarrow \log _{20}20<\log _{20}40<\log _{20}20^2$
$\Rightarrow 1< \log _{20}40<2 \quad [\because \log a^m=m\log a, \log _aa=1]$
$\Rightarrow 1<-y<2\Rightarrow -2<y<-1$

The value of $\displaystyle \log _{\frac{2}{3}}\frac{5}{6}$ is

  1. less than zero.

  2. greater than zero and less than one.

  3. greater than one.

  4. none of these


Correct Option: B
Explanation:

$\displaystyle \frac { 2 }{ 3 } <\frac { 5 }{ 6 } <1$
Taking log with base $\displaystyle \frac { 2 }{ 3 } $
$\displaystyle \log _{ \frac { 2 }{ 3 }  }{ \frac { 2 }{ 3 }  } >\log _{ \frac { 2 }{ 3 }  }{ \frac { 5 }{ 6 }  } >\log _{ \frac { 2 }{ 3 }  }{ 1 } $  (Since the base $\displaystyle \frac { 2 }{ 3 } <1)$
$\displaystyle \Rightarrow 1>\log _{ \frac { 2 }{ 3 }  }{ \frac { 5 }{ 6 }  } >0$

Value of $\displaystyle \log _{4}18 $ is:

  1. an irrational number

  2. a rational number

  3. natural number

  4. whole number


Correct Option: A
Explanation:
$\log _{4}{18}=\log _{2^{2}}{(2.3^{2})}=\dfrac{1}{2}\log _{2}(2.3^{2})$
$=\dfrac{1}{2}\left [ \log _{2}{2}+\log _{2}{3^{2}} \right ]=\dfrac{1}{2}\left [ 1+2.\log _{2}{3} \right ]$

$\log _{2}{3}$ is an irrational number.

Hence, $\dfrac{1}{2}\left [ 1+2.\log _{2}{3} \right ]$ is also an irrational number.

$\log _4 $1 is equal to

  1. $1$

  2. $0$

  3. $\infty$

  4. none of these


Correct Option: B
Explanation:

$log _a 1 = m$
$\Rightarrow a^m = 1 a^0    \Rightarrow m = 0$

If $x=\log _{ a }{ bc } ,y=\log _{ b }{ ca } ,z=\log _{ c }{ ab } $, then the value of $\dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $ will be

  1. $x+y+z$

  2. $1$

  3. $ab+bc+ca$

  4. $abc$


Correct Option: B
Explanation:

Now, $1+x=\log _{ a }{ a } +\log _{ a }{ bc } =\log _{ a }{ abc } $
$\Rightarrow \dfrac { 1 }{ 1+x } =\log _{ abc }{ a } $
Similarly, $\dfrac { 1 }{ 1+y } =\log _{ abc }{ b } $ and $\dfrac { 1 }{ 1+z } =\log _{ abc }{ c } $
$\therefore \dfrac { 1 }{ 1+x } +\dfrac { 1 }{ 1+y } +\dfrac { 1 }{ 1+z } $
                    $=\log _{ abc }{ a } +\log _{ abc }{ b } +\log _{ abc }{ c } $
                    $=\log _{ abc }{ abc } =1$

The characteristic of a number having $m$ $(m>1)$ digits is given by,

  1. $m-1$

  2. $m+1$

  3. $m$

  4. None of the above


Correct Option: A
Explanation:

If a number is $N>0$, then $\log _{10}N$ will have two parts, the integral part is known the characteristic and the decimal part is known as mantissa.

$2$ digit belongs from $[10,100]$ where $\log _{10}10=1$ and $\log _{10}100=2$
Similarly, $m$ digit number belongs from $[10^{m-1},10^m]$, where $\log _{10}10^{m-1}=m-1$ and $\log _{10}10^m=m$.
Thus any number between $[10^{m-1},10^m]$ will have $m-1$ as the integral part.
Thus the characteristic of a number having $m$ digits is given by $m-1$.

Calculate $x$, to the nearest tenth: $\log _{12} 640 = x$

  1. $1.7$

  2. $2.6$

  3. $2.8$

  4. $53.3$

  5. $7,680$


Correct Option: B
Explanation:

Given, $x= \log _{ 12 }{ 640 } $

$=\log _{ 12 }{ 144\times 4.44 } $
$=\log _{ 12 }{ 144 } +\log _{ 12 }{ 4.44 } $
$=2+\log _{ 12 }{ 4.44 }$
$= 2.6$

If $\displaystyle { log } _{ 5 }{ log } _{ 5 }{ log } _{ 2 }x=0$, then the value of $x$ is

  1. $32$

  2. $125$

  3. $625$

  4. $25$


Correct Option: A
Explanation:

Given, $\displaystyle { log } _{ 5 }{ log } _{ 5 }{ log } _{ 2 }x=0$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=50$
$\displaystyle \Rightarrow \quad { log } _{ 5 }{ log } _{ 2 }x=1\Rightarrow { log } _{ 2 }x=5$
$\displaystyle \Rightarrow x={ 2 }^{ 5 }=32$

The value of $\log _{10} 0.0006024$ is equal to

  1. $\overline {3}.7979$

  2. $\overline {1}.9779$

  3. $\overline {4}.7799$

  4. $0.7279$


Correct Option: C
Explanation:

$\log _{10}0.0006024$

Characteristics$=-4$
For mantissa ,read from the table $6024.$
Mantissa$=7799$ 
Thus, $\log _{10}0.0006024=$ characteristics of $0.0006024+$ mantissa of $0.0006024$
$=-4+0.7799$
$=\bar4.7799$
Hence, C is the correct option.

Using logarithm table, determine the value of $\log _{10}0.5432$.

  1. $\overline {1}.7350$

  2. $\overline {2}.7350$

  3. $0.7350$

  4. $0.07350$


Correct Option: A
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.

Antilog of the number $( -8.654)$ is equal to

  1. $2.18\times 10^{-8}$

  2. $2.18\times 10^{-9}$

  3. $2.218\times 10^{-9}$

  4. $2.218\times 10^{-8}$


Correct Option: C
Explanation:
$(-8.654)=-8+(-0.654)$
$=-8+(-0.654)+1-1$
$=-9+(1-0.654)=-9+0.346$

Actual digits of 0.346 from the log table $=2.218$

$\therefore$ $ anti \log(-8.654)=2.218\times 10^{-9}$

The antilog of $\overline {1}.8840$is equal to

  1. $76.56$

  2. $\overline {1}.7656$

  3. $0.7656$

  4. $7.656$


Correct Option: C
Explanation:

Antilog $\bar1.8840$

Characteristics$=-1$
Value of $0.8840$ from antilog table $=7656$

Now number of zeroes after decimal will be $0.$
Antilog $\bar1.8840=0.7656$
Hence, C is the correct option.

The antilog of the number $2.9586$ is equal to

  1. $909.1$

  2. $90.91$

  3. $9.091$

  4. $0.9091$


Correct Option: A
Explanation:

The number before decimal point is $2,$ so decimal point will be after $3$ digits.

Value of $0.9586$ from antilog table $=9078+12=9090$
Now place a decimal point after $3$ digits of the number from left we get, $909.0$
Antilog $2.9586=909.0$
Hence, A is the correct option.

The antilog of $(1.32)$ is equal to

  1. $20$

  2. $20.98$

  3. $20.89$

  4. $20.79$


Correct Option: C
Explanation:

The number before decimal point is $1,$ so decimal point will be after $2$ digits.

Value of $1.32$ from antilog table $=2089$
Now place a decimal point after $2$ digits of the number from left we get, $20.89$
Antilog $1.32=20.89$
Hence, C is the correct option.

The antilog of the number $0.2015$ is equal to

  1. $\overline {1}.1591$

  2. $1.591$

  3. $0.01591$

  4. $15.91$


Correct Option: B
Explanation:

The number before decimal point is $0,$ so decimal point will be after $1$ digits.

Value of $0.2015$ from antilog table $=1589+2+=1591$
Now place a decimal point after $1$ digits of the number from left we get, $1.591$
Antilog $0.2015=1.591$
Hence, B is the correct option.

If there are $n$ zeros after the decimal point, then the characteristic of that number will be

  1. $n+1$

  2. $-n+1$

  3. $-(n+1)$

  4. $n-1$


Correct Option: C
Explanation:

For number less than $1,$ if there are $n$ zeroes after the decimal point, then the characteristics of that number will be $-(n+1).$

and For number  greater than $1,$ if  there are $n$ number of zeroes are on the left sides of the digits then characteristics will be $(n+1).$
Hence, C is the correct option.

Evaluate using log tables: $\sqrt [3] {\dfrac {16.23}{426.8}}$

  1. $0.6332$

  2. $0.3632$

  3. $0.3362$

  4. $0.3624$


Correct Option: C
Explanation:

$\log { y } =\cfrac { 1 }{ 3 } \left[ \log { (16.23) } -\log { (426.8) }  \right] \ \log { y } =\cfrac { 1 }{ 3 } \left[ 1.2103-2.63022 \right] \ \log { y } =-0.4733\ y={ 10 }^{ -0.4733 }=0.3362$

If $f(x) = \log x$, then $f^{-1}x  $ is

  1. $\log\dfrac1x$

  2. $\log x^2$

  3. $anti\log(x)$

  4. $anti\log\dfrac1x$


Correct Option: C
Explanation:

Let,$f(x)=y=\log x$

$\Rightarrow f(x)=y=\log x$
$\Rightarrow $ anti$\log (y)=x$
Therefore, Inverse of $f(x)$ i.e. $f^{-1}(x)=$anti$\log (x)$
Hence, C is the correct option.

The antilog of $(4.8779)$ will be 

  1. $7500$

  2. $750$

  3. $75500$

  4. $750000$


Correct Option: C
Explanation:

The number before decimal point is $4,$ so decimal point will be after $5$ digits.

Value of $0.8779$ from antilog table $=7534+16=7550$
Now place a decimal point after $5$ digits of the number from left we get, $75500$
Antilog $4.8779=75500$

The value of $\log _{10} 8$ is equal to

  1. $.903$

  2. $3.901$

  3. $.301$

  4. None of the above


Correct Option: A
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.

Find the value of $\dfrac {\log _{10} 72}{\log _{10} 8}$ using log table

  1. $\log _{10} 9$

  2. $1+\dfrac{.954}{.903}$

  3. $2$

  4. $\dfrac{.903+.954}{.954}$


Correct Option: B
Explanation:
$\dfrac{\log _{10}{72}}{\log _{10}{8}}=\dfrac{\log _{10}{(8\times9)}}{\log _{10}{8}}=\dfrac{\log _{10}{8}+\log _{10}{9}}{\log _{10}{8}}$
$=1+\dfrac{\log _{10}{3^{2}}}{\log _{10}{2^{3}}}=1+\dfrac{2\log _{10}{3}}{3\log _{10}{2}}$
$=1+\dfrac{2.(0.477)}{3.(0.301)}=1+\dfrac{(0.954)}{(0.903)}$

Find the value of $\log _{10} 72$ using log table

  1. $0.901+0.909$

  2. $0.903+0.954$

  3. $1.890$

  4. $2.104$


Correct Option: B
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 $AntiLog(.2817)$.

  1. $1.70$

  2. $2.19$

  3. $1.91$

  4. $1.99$


Correct Option: C
Explanation:

We know that $AntiLog(x)=10^{x}$

$\Rightarrow AntiLog(0.2817)=10^{0.2817}=1.91$
This exact value should be find by using anti logarithm table.
Therefore the correct option is $C$

Find the value of $\log _{10} {\dfrac{64^{2.1}\times 81^{4.2}}{49^{3.4}}}$ using log table

  1. $2.1 \times 6 \times .303+ 4.2 \times 2 \times .854- 3.4 \times 2 \times .745$

  2. $2.1 \times .303+ 4.2 \times .954- 3.4 \times .845$

  3. $2.1 \times 6 \times .303- 4.2 \times 2 \times .954+3.4 \times 2 \times .845$

  4. $2.1 \times 6 \times .303+ 4.2 \times 2 \times .954- 3.4 \times 2 \times .845$


Correct Option: D
Explanation:
$\log _{10}{64^{2.1}}+\log _{10}{81^{4.2}}-\log _{10}{49^{3.4}}$
$= 2.1\log _{10}{64}+4.2\log _{10}{81}-3.4\log _{10}{49}$
$=2.1\log _{10}{2^{6}}+4.2\log _{10}{3^{4}}-3.4\log _{10}{7^{2}}$
$=2.1\times6\times(0.303)+4.2\times4\times(0.477)-3.4\times2\times(0.845)$
$=2.1\times6\times(0.303)+4.2\times2\times(0.954)-3.4\times2\times(0.845)$

Let $x = (0.15)^{20}$. Find the characteristic in the logarithm of $x$ to the base $10$.

  1. $17$

  2. $21$

  3. $-21$

  4. $-17$


Correct Option: D
Explanation:

Given $x=(0.15)^{20}$

By applying $\log$ on both sides , we get $\log _{ 10 }{ x } =20\log _{ 10 }{ (0.15) } =-16.478$
$\Rightarrow \log _{ 10 }{ x } =-17+0.5218$
The integral part of $\log _{ 10 }{ x } $ is called characteristic
Therefore the characteristic of given number is $-17$
So option $D$ is correct

Find the value of ${\log _{10} 72} + {\log _{10} {\dfrac{1}{8}}}$ using log table

  1. $0.903$

  2. $0.303$

  3. $0.954$

  4. $1.234$


Correct Option: C
Explanation:
$\log _{10}{72}+\log _{10}\left (\dfrac{1}{8}  \right )$
$=\: \log _{10}{\left (72\times \dfrac{1}{8}  \right )}=\log _{10}{9}=\log _{10}{3^{2}}$
$=\: 2.\log _{10}{3}=2(0.477)=0.954$

If $x^2+y^2=25$ , then $log _5 \begin {bmatrix} Max (3x+4y) \end {bmatrix}$ is

  1. $2$

  2. $3$

  3. $4$

  4. $5$


Correct Option: A
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 $x=3 \,and\, y=4$

$S=3x+4y=3(3)+4(4)=25$
$log _5 (3x+4y)=log _5(s)=log _5(25)=log _5(5^2)=2$

If the mantissa of $\log 2125 =3.3275$, find the mantissa of $\log21.25$

  1. $1.3273$

  2. $2.3273$

  3. $0.3273$

  4. $32.2321$


Correct Option: A
Explanation:

Given that:

$\log2125=3.3273$
Now, $\log21.25=\log\cfrac{2125}{100}$
$=\log 2125-\log100$
$=3.3273-\log10^2$
$=3.3273-2$
$=1.3273$

The logarithm of $0.0625$ to the base $2$ is:

  1. $0.025$

  2. $0.25$

  3. $5$

  4. $-4$

  5. $-2$


Correct Option: D
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 number of zeros between the decimal point and first significant digit of ${\left(0.036\right)}^{16}$ where $log2=0.301$ and $log3=0.477$

  1. $21$

  2. $22$

  3. $23$

  4. $24$


Correct Option: C
Explanation:

$y=(0.036)^{16}$

$\Rightarrow \log (y)= 16 log(0.036)$
$\Rightarrow \log (y)= 16 \log(\cfrac{36}{1000})=16[\log 36- 3\log _{10}{10}]$
$\Rightarrow \log(y)= 16[\log (2^2\times 3^2)-3]=16[2(\log 2+\log 3)-3]$
$\Rightarrow \log(y)=16[2(0.301+0.477)-3]=-23.104<-23$
So, that number has $23$ zeros after this point.

Given $log _{10}2=a$ and $log _{10}3=b$, if $3x+2=25$, the value of x in terms of $a$ and $b$ is $x=(10^{k}+1)$. K=?

  1. $\dfrac{a-1}{b}$

  2. $a-b+1$

  3. $\dfrac{1+a}{b}$

  4. $\dfrac{b}{1-a}$


Correct Option: B
Explanation:

$\log _{10}{2}=a$

$\Rightarrow 2=(10^a)$
$\log _{10}{3}=b$
$\Rightarrow 3=(10^b)$
$3x+2=25$
$x=\cfrac{23}{3}=(2\times \cfrac{10}{3}+1)=10\cfrac{(10)^a}{(10)^b}+1$
$x=10^{(a+1-b)}+1$
$x=(10^{a-b+1}+1)$

Let $N=\dfrac{\log _{3}135}{\log _{15}3}-\dfrac{\log _{3}5}{\log{405}3}$, then $N$ is

  1. a natural number

  2. a prime number

  3. a rational number

  4. an integer


Correct Option: A,B,C,D

If $x=198!$ then value of the expression $\dfrac {1}{\log _{2}x}+\dfrac {3}{\log _{2}x}+...\dfrac {198}{\log _{2}x}$ equals ?

  1. $-1$

  2. $0$

  3. $1$

  4. $198$


Correct Option: C

The value of $\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}$ is

  1. $3$

  2. $0$

  3. $2$

  4. $1$


Correct Option: A
Explanation:
Consider
$\dfrac{log _2 24}{log _{96} 2}-\dfrac{log _2192}{log _{12}{2}}\\$
$=\dfrac{log24.log96-log192log12}{(log2)^2}$
$=\dfrac{log(2^3 \times 3)log(2^5\times 3)-log(2^6\times3)log(2^2\times3)}{(log2)^2}$
$=\dfrac{(3log2+log3)(5log2+log3)-(6log2+log3)(2log2+log3)}{(log2)^2}$
$=\dfrac{15(log2)^2-12(log2)^2}{(log2)^2}$
$=3\dfrac{(log2)}{log2}$
$=3$
Option A is the correct answer.

The greatest value of $(4\log _{10}{x}-\log _{2}{(0.0001)})$ for $0 < x < 1$ is

  1. $4$

  2. $-4$

  3. $8$

  4. $-8$


Correct Option: A

If $P$ is the number of natural numbers whose logarithm to the base $10$ have the characteristic $p$ and $Q$ is the number of natural numbers logarithm of whose reciprocals to the base $10$ have the characteristic $-q$, then find the value of $\log _{10}P-\log _{10}Q$.

  1. $p-q+1$

  2. $p+q-1$

  3. $p+q$

  4. $p-q$


Correct Option: A
Explanation:

Let $x$ and $y$ be the numbers whose logarithm to the base $10$ have the characteristic $p$ and $q$ respectively.
$10^{p}\leq x< 10^{p+1}\Rightarrow P=10^{p+1}-10^{p}\Rightarrow P=9\times 10^{p}$
Similarly, $10^{q-1}< y\leq 10^{q}$
$\Rightarrow $   $Q=10^{q}-10^{q-1}=10^{q-1}\left ( 10-1 \right )=9\times 10^{q-1}$
$\therefore $   $\log _{10}P-\log _{10}Q=\log _{10}\left ( P/Q \right )=\log _{10}10^{p-q+1}=p-q+1$

Find the number of positive integers which have the characteristic $3$, when the base of the logarithm is $7$.

  1. $2058$

  2. $1029$

  3. $1030$

  4. $2060$


Correct Option: A
Explanation:

Let there be $N$ integers whose characteristic is 3, when base of log is 7
Then, $\log _{ 7 }{ N } =x$ where $3\le x<4$
As $3\le x<4$
$3\le \log _{ 7 }{ N } <4\ \Rightarrow { 7 }^{ 3 }\le N<{ 7 }^{ 4 }\ \Rightarrow N={ 7 }^{ 4 }-{ 7 }^{ 3 }\ \Rightarrow N=2401-343=2058$ 

The value of $\displaystyle anti\log _5\left [\frac {\tan^2\left (\frac {\pi}{5}\right )+\tan^2\left (\frac {2\pi}{5}\right )+20}{\cot^2\left (\frac {\pi}{5}\right )+\cot^2\left (\frac {2\pi}{5}\right )+28}\right ]$ is equal to

  1. odd number

  2. even number

  3. prime number

  4. composite number


Correct Option: A,C
Explanation:
$\frac { { tan }^{ 2 }(\frac { \Pi  }{ 5 } )+{ tan }^{ 2 }(\frac { 2\Pi  }{ 5 } )+20 }{ { cot }^{ 2 }(\frac { \Pi  }{ 5 } )+{ cot }^{ 2 }(\frac { 2\Pi  }{ 5 } )+28 } \\ =\frac { { (\sqrt { 5-2\sqrt { 5 }  } ) }^{ 2 }+(\sqrt { \frac { 5 }{ 5-2\sqrt { 5 }  }  } )^{ 2 }+20 }{ { (\frac { 1 }{ \sqrt { 5-2\sqrt { 5 }  }  } ) }^{ 2 }+{ (\sqrt { \frac { 5-2\sqrt { 5 }  }{ 5 } ) }  }^{ 2 }+28 } \\ =\frac { 5-2\sqrt { 5 } +\frac { 5 }{ 5-2\sqrt { 5 }  } +20 }{ \frac { 1 }{ 5-2\sqrt { 5 }  } +\frac { 5-2\sqrt { 5 }  }{ 5 } +28 } \\ =\frac { 5[(5-{ 2\sqrt { 5 } ) }^{ 2 }+5+20(5-2\sqrt { 5 } )] }{ 5+(5-{ 2\sqrt { 5 } ) }^{ 2 }+140(5-2\sqrt { 5 } ) } \\ =\frac { 5(45-20\sqrt { 5 } +5+100-40\sqrt { 5 } ) }{ 5+45-20\sqrt { 5 } +700-200\sqrt { 5 }  } \\ =\frac { 750-300\sqrt { 5 }  }{ 750-300\sqrt { 5 }  } \\ antilog _{ 5 }[\frac { { tan }^{ 2 }(\frac { \Pi  }{ 5 } +{ tan }^{ 2 }(\frac { 2\Pi  }{ 5 } )+20 }{ { cot }^{ 2 }(\frac { \Pi  }{ 5 } )+{ cot }^{ 2 }(\frac { 2\Pi  }{ 5 } )+28 } ]={ 5 }^{ 1 }\quad =5$

Evaluate using logarithm table: $\dfrac {28.45 \times \sqrt [3] {0.3254}}{32.43 \times \sqrt [5] {0.3046}}$

  1. $0.7666$

  2. $0.7656$

  3. $0.5686$

  4. $0.2936$


Correct Option: B
Explanation:

Let $y=\dfrac { 28.45\times \sqrt [ 3 ]{ .3254 }  }{ 32.43\times \sqrt [ 3 ]{ .3046 }  } $

$ \ln { y } =\ln { 28.45 } +\ln { \sqrt [ 3 ]{ .3254 }  } -(\ln { 32.43 } +\ln { \sqrt [ 5 ]{ .3046 }  } )\ \ln { y } =\ln { 25.45 } +\dfrac { 1 }{ 3 } \ln { .3245- } \ln { 32.43 } -\dfrac { 1 }{ 5 } \ln { .4046 } \ \ln { y } =3.236+(-.375)-3.479-(-.237)\ \ln { y } =-.381$
$ y=$ anti $\ln { (-.381) } $

$ y=.7656$
So, option B is correct.

If $\log _{10} 2 = 0.3010$, then the number of digits in $2^{64}$ is

  1. $18$

  2. $24$

  3. $22$

  4. $20$


Correct Option: D
Explanation:

Given $\log _{ 10 }{ 2 } =0.301$

$\log _{ 10 }{ 2^{64} } =64 \times \log _{ 10 }{ 2 } =64 \times 0.3010=19.264$
$\Rightarrow 2^{64}=10^{19.264}$
The number of digits in $10^{19}$ is $20$ , there will be $21$ digits from $10^{21}$
The number $10^{19.264}$ lies between them
Therefore the number of digits in $10^{19.264}$ is $20$
Therefore the correct option is $D$

If $\log _{10} 3 = 0.4771$, then the number of zeros after the decimal in $3^{-100}$ is

  1. $47$

  2. $48$

  3. $49$

  4. $50$


Correct Option: A
Explanation:

The number of zeroes will be
$=|log _{10}(3^{-100})|$
$=|-100(log _{10}(3))|$
$=|-47.71|$
$=47.71$
Taking the integral part (since number of zeroes has to be an integer, there will be $47$ zeros.

Approximate of $\log _{11}21$ is

  1. 1.27

  2. 1.21

  3. 1.18

  4. 1.15

  5. 1.02


Correct Option: A
Explanation:

Approximate value of $\log _{ 11 }{ 21 } $

$=\log _{ 11 }{ (7\times 3) } $
$=\log _{ 11 }{ 7 } +\log _{ 11 }{ 3 }$
$ =0.8115+0.4581$
$=1.27$

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