Tag: logarithm and its uses

Questions Related to logarithm and its uses

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:

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