Tag: more about logarithms

Questions Related to more about logarithms

Which is the correct order for a given number $\alpha$ in increasing order.

  1. $\log _{2} \alpha, \log _{e} \alpha, \log _{3} \alpha, \log _{10} \alpha$

  2. $\log _{10} \alpha, \log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha$

  3. $\log _{10} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{3} \alpha$

  4. $\log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{10} \alpha$


Correct Option: A
Explanation:
We know that, greater the value of base lesser is the value of the logarithm

Also, $10<3<e(2.71)<2$

$\therefore $ For any positive number $\alpha $,

${ log } _{ 2 }\alpha >{ log } _{ e }\alpha >{ log } _{ 3 }\alpha >{ log } _{ 10 }\alpha $

If $\log _3{(\log _3{a})}+\log _{\cfrac{1}{3}}{\left(\log _{\cfrac{1}{3}}{b}\right)}=1$, then the value of $ab^3$ is 

  1. $9$

  2. $3$

  3. $1$

  4. $\cfrac{1}{3}$


Correct Option: C

The value of $\log _a n\times\log _n m $  is equal to

  1. $\log _a m$

  2. $\log _m a$

  3. $\dfrac{\ln m}{\ln a}$

  4. $\dfrac{\ln a}{\ln m}$


Correct Option: A,C
Explanation:

$\because \displaystyle \log _{ n }{ m } =\frac { \log{ m }  }{ \log { n }  } $
$\therefore \log _an\times \log _nm=\dfrac{\log n}{\log a}\times \dfrac{\log m}{\log n}=\dfrac{\log m}{\log a}=\log _am$

If $\displaystyle \log _{10}\left [ \log _{10}\left ( \log _{10}x \right ) \right ]=0 $

  1. x = $\displaystyle 10^{3}$

  2. x = $\displaystyle 10^{10}$

  3. x = $\displaystyle 15^{5}$

  4. None


Correct Option: B
Explanation:

 $ \log _{10}\left [ \log _{10}\left ( \log _{10}x \right ) \right ]=0 $
$log _{10} log _{10} x=1$
$log _{10 }x=10$
$x=10^{10}$

If $x = \displaystyle \frac{y}{(1 + x)^p}$, then $p$ is equal to

  1. $\displaystyle \frac{\displaystyle \log _e \left ( \frac{y}{x} \right )}{\log _e (1 + a)}$

  2. $\log \displaystyle \left { \frac{y}{x(1+ a)} \right }$

  3. $\log \displaystyle \left { \frac{y - x}{1+ a} \right }$

  4. $\displaystyle \frac{\log y}{\log { x(1 + a) }}$


Correct Option: A
Explanation:

Since $x = \displaystyle \frac{y}{(1 + a)^p}$


$\therefore   (1 + a)^p = \displaystyle \frac{y}{x}$

or $p   \log _e (1 + a) = \log _e\dfrac{y}{x}$

or $\displaystyle p = \dfrac{\displaystyle log _e\left ( \frac{y}{x} \right )}{log _e (1 + a)}$

If $\log _{ 5 }{ x } =y$, then ${5}^{5y}$ is

  1. $\cfrac{x}{5}$

  2. $5x$

  3. $\log _{ x }{ 5 } $

  4. ${x}^{5}$


Correct Option: D
Explanation:

$\log _5 x =y $

$5^{\log _5 x }=5^y $
$5^y =x$

Then , $5^{5y} = (5^y)^5 = x^5$

The value of $\log _{ 2 }{ 7 } $ is:

  1. an integer

  2. a prime number

  3. a rational number

  4. an irrational number


Correct Option: D
Explanation:
Suppose $\log _{2}{7}$ is rational.

$\Rightarrow \: \log _{2}{7}=\dfrac{a}{b}\:\Rightarrow \: 7=2^{a/b}$
$\Rightarrow \: 7^{b}=2^{a}$

But $2^{a}$ is even and $7^{b}$ is odd.
Hence, our assumption is wrong.

$\Rightarrow \: \log _{2}{7}$ is irrational.

The value of $x$ satisfying $\log _{ 243 }{ x } =0.8$

  1. $81$

  2. $1.8$

  3. $2.43$

  4. $27$


Correct Option: A
Explanation:

$\log _{243}x=0.8$

$\Rightarrow x=243^{0.8}$
$\Rightarrow x=81$
Hence, A is the correct option.

If $\log _{ x }{ \left( 7x-10 \right)  } =2$, then find the value(s) of $x$.

  1. $2$

  2. $3$

  3. $4$

  4. $5$


Correct Option: A,D
Explanation:

$\log _x(7x-10) =2$

$\Rightarrow x^2 =7x-10$ ...... $[\because a^{\log _{a} x}=x]$

$\Rightarrow x^2-7x+10=0$

$\Rightarrow x^2-5x-2x-10 =0$

$\Rightarrow (x-5)(x-2)=0$

Hence, $x =5 \; or \; x =2$

if $y=\left( \log _{ 2 }{ 3 }  \right) \left( \log _{ 3 }{ 4 }  \right) ....\left( \log _{ 31 }{ 32 }  \right) $, then

  1. $4< y\le 5$

  2. $y=5$

  3. $4< y< 6$

  4. $y=6$


Correct Option: A,B,C
Explanation:

Let $y = (\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 )$

Then, $2^y =2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

By laws of exponents and the definition of a logarithm,

$2^{((\log _2 3 ) (\log _3 4 ) ( \log _4 5 ) … (log _{31} 32 ))} $

$=(2^{(\log _23)})^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$= 3^{((\log _3 4 ) ( \log _4 5 ) … (\log _{31} 32 ))} $

$=(3^{(\log _34)})^{(( \log _4 5 ) … (\log _{31} 32 ))} $

$=4^{(( \log _4 5 ) … (\log _{31} 32 ))} $ .......

$=31^{(\log _{31}32)} =32$

$\therefore 2^y =32$

$y=5$