Tag: a relation between logarithmic functions

Questions Related to a relation between logarithmic functions

$\log _a {bc}= x, \log _b {ac}= y , \log _c {ab}= z$, then $\dfrac{1}{x + 1} + \dfrac{1}{y + 1} + \dfrac{1}{z + 1} = $

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $0$

  2. $1$

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

  4. none

Show answer
Correct Option: B
Explanation:

We have,

$x=\log _a bc$

$x=\dfrac{\log bc}{\log a}$                     $\because \log _a m=\dfrac{\log m}{\log a}$


Similarly,

$y=\dfrac{\log ac}{\log b}$

$z=\dfrac{\log ab}{\log c}$


Since,

$\Rightarrow \dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}$


Therefore,

$\Rightarrow \dfrac{1}{\dfrac{\log bc}{\log a}+1}+\dfrac{1}{\dfrac{\log ac}{\log b}+1}+\dfrac{1}{\dfrac{\log ab}{\log c}+1}$

$\Rightarrow \dfrac{\log a}{{\log bc}+{\log a}}+\dfrac{\log b}{{\log ac}+{\log b}}+\dfrac{\log c}{{\log ab}+{\log c}}$

$\Rightarrow \dfrac{\log a}{{\log abc}}+\dfrac{\log b}{{\log abc}}+\dfrac{\log c}{{\log abc}}$

$\Rightarrow \dfrac{\log a+\log b+\log c}{{\log abc}}$

$\Rightarrow \dfrac{\log abc}{{\log abc}}$

$\Rightarrow 1$


Hence, this is the answer.


The remainder when ${75^{{{75}^{75}}}}$ is divided by $37$.

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $0$

  2. $1$

  3. $3$

  4. can't be determine

Show answer
Correct Option: B
Explanation:
Since $75 ≅ 1 modulo 37$

$75 \times 75 ≅ 1 *1 (=1) modulo 37$

$75 \times 75\times 75 ≅ 1*1*1 (=1) modulo 37$

So $75 ^{\ any \ integer}≅ $1$ modulo $37$
So$\ {75^{{75}^{75}}}$ ≅ $1 $modulo $37$

The remainder is 1.
*The digits (=1) are the remainders when divided by 37.

If $ 3^{\log _{4}{x}}=27$, then $x$ is equal to

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $16$

  2. $64$

  3. $27$

  4. $\log _2 16$

Show answer
Correct Option: B
Explanation:

$ 3^{\log _{4}{x}}=27$
$\Rightarrow { 3 }^{ \log _{ 4 }{ x }  }={ 3 }^{ 3 }$
Equating powers of $3$
$\log _{ 4 }{ x } =3$
Taking exponential of the above equation, we get
${ 4 }^{ 3 }=x$
$\therefore x=64$

The value of $3^{\log _{ 4 }{ 5 }} -5 ^{\log _{ 4 }{ 3 }}$ is

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $0$

  2. $1$

  3. $2$

  4. none of these

Show answer
Correct Option: A
Explanation:

We have,
${ 3 }^{ \log _{ 4 }{ 5 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  }$
${ =5 }^{ \log _{ 4 }{ 3 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  },$ ($\because { x }^{ \log _{ a }{ y }  }={ y }^{ \log _{ a }{ x }  }$)
$=0$

If $\log _{k}x.\log _{5}k=\log _{x}5,k\neq 1,k> 0$, then the value of $x$ is equal to

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $k$

  2. $\displaystyle \frac{1}{5}$

  3. $5$

  4. none of these

Show answer
Correct Option: B,C
Explanation:

$\log _{ k }{ x } \log _{ 5 }{ k } =\log _{ x }{ 5 } \ \Rightarrow \dfrac { \log { x } \log { k }  }{ \log { k } \log { 5 }  } \left[ \because \log _ba=\dfrac{\log a}{\log b}\right]=\dfrac { \log { 5 }  }{ \log { x }  } \ \Rightarrow { \left( \log { x }  \right)  }^{ 2 }={ \left( \log { 5 }  \right)  }^{ 2 }\ \Rightarrow \log { x } =\pm \log { 5 }=\log 5 , \log 5^{-1}, \left[n\log a=\log a^n \right] \ \therefore \quad x=5,\dfrac { 1 }{ 5 } $

Ans: B.C

${ \log } _{ a }{ x }^{ n }=n{ \log } _{ a }x$

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. True

  2. False

Show answer
Correct Option: A
Explanation:
Step 1:
Let $m = \log _a x$

Step 2: Write in exponent form
$x = a^m$

Step 3: Raise both sides to the power of $n$
$x^n = ( a^m )^n$

Step 4: Convert back to a logarithmic equation
$\log _a x^n = mn$

Step 5: Substitute for $m = \log _a x$
$\log _a x^n = n \log _a x$

If $\displaystyle 5x^{log _23} + 3^{log _2x} = 162$ then logarithm of $x$ to the base 4 has the value equal to :

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $2$

  2. $1$

  3. $-1$

  4. $3/2$

Show answer
Correct Option: D
Explanation:
Given,

$5x^{\log _2\left(3\right)}+3^{\log _2\left(x\right)}=162$

$3^{\log _2\left(x\right)}=162-5x^{\log _2\left(3\right)}$

$\Rightarrow \log _2\left(3^{\log _2\left(x\right)}\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$\Rightarrow \log _2\left(x\right)\log _2\left(3\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$\Rightarrow \log _2\left(x^{\log _2\left(3\right)}\right)=\log _2\left(162-5x^{\log _2\left(3\right)}\right)$

$x^{\log _2\left(3\right)}=162-5x^{\log _2\left(3\right)}$

$x=27^{\frac{665}{1054}}\approx 8$

Now,

$\log _4x$

$=\log _4 8$

$=\dfrac{3}{2}$

The value of $ a^{\frac{\log _b (\log _b N)}{\log _b a}}$ is

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $\log _b (N-b)$

  2. $\log _b (N+b)$

  3. $\log _b\dfrac Nb$

  4. $\log _b N$

Show answer
Correct Option: D
Explanation:
Using $\log _p q =\dfrac{\log q}{\log p}$
$\displaystyle= a^{\cfrac{\log _b (\log _b N)}{\log _b a}}$
$\displaystyle = a^{\cfrac{\log(\log _b N)/\log b}{\log a/\log b}}$
$=\displaystyle a^{\displaystyle \log _a (\log _b N)}$
$=\log _b N[\because a^{\log _ab=b}]$


If ${ log } _{ 4 }5=a\quad and\quad { log } _{ 5 }6=b,\quad then\quad { log } _{ 3 }2$ is equal to

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $\dfrac { 1 }{ 2a+1 } $

  2. $\dfrac { 1 }{ 2b+1 } $

  3. $2ab+1$

  4. $\dfrac { 1 }{ 2ab-1 } $

Show answer
Correct Option: D
Explanation:
Given,

$\log _{4}5=a$

$\log _{5}6=b$

now,

$ab=\log _{4}5 \times \log _{5}6$

$=\dfrac{\log 5}{\log 4} \times \dfrac{\log 6}{\log 5}$

$=\dfrac{\log 6}{\log 4}$

$ab=\log _4{6}$

$=\dfrac{1}{2}\log _2 6$

$=\dfrac{1}{2}\log _2 (2\times 3)$

$=\dfrac{1}{2}(\log _2 2+\log _2 3)$

$=\dfrac{1}{2}(1+\log _2 3)$

$2ab=1+\log _2 3$

$\log _2 3=2ab-1$

$\therefore \log _3 2=\dfrac{1}{2ab-1}$

If $4^{\log _{2}\log x}=\log x-\left ( \log x \right )^{2}+1$ (base is e), then find the value of $x$

maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

  1. $x=e$

  2. $x=2e$

  3. $x=3e$

  4. none of these

Show answer
Correct Option: A
Explanation:

Given equation is 
$4^{\log _{2}\log x}=\log x-\left ( \log x \right )^{2}+1$

Now, $\log _{2}\log x$ is meaningful if $\log x > 0$.
Since $4^{\log _{2}\log x}=2^{2\log _{2}\log x}=\left ( 2^{\log _{2}\log x} \right )^{2}=\left ( (\log x)^{\log _{2}2} \right )^{2}[\because a^{\log _bc}=c^{\log _ba}]$
             $=\left ( \log x \right )^{2}$    $ \left ( \because a^{\log _{a}x}=x, a> 0, a\neq 1 \right )$

So the given equation reduces to
$2\left ( \log x \right )^{2}-\log x-1=0$.
$\displaystyle \Rightarrow \log x=1, \log x=-\frac{1}{2}$.
But $\log x> 0$
Hence,    $\log x=1$, i.e.,$x=e$