Tag: properties of logarithms

Questions Related to properties of logarithms

The value of $\left( \log _{ b }{ a }  \right) \left( \log _{ c }{ b }  \right) \left( \log _{ a }{ c }  \right) $ is equal to

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

  1. $0$

  2. $\log { abc } $

  3. $1$

  4. $10$

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Correct Option: C
Explanation:

Let $x=\left( \log _{ b }{ a }  \right) \left( \log _{ c }{ b }  \right) \left( \log _{ a }{ c }  \right) \ \Rightarrow x=\left( \dfrac { \log { a }  }{ \log { b }  }  \right) \left( \dfrac { \log { b }  }{ \log { c }  }  \right) \left( \dfrac { \log { c }  }{ \log { a }  }  \right),\left[\because \log _yx=\cfrac{\log x}{\log y}\right] \ \Rightarrow x=1$

Ans: C

Sometimes to solve an equation, we may use the identity ${ a }^{ \log _{ a }{ b }  }=b,b>0,a>0,a\neq 1$
Then solution set of $3{ x }^{ \log _{ 5 }{ 2 }  }+{ 2 }^{ \log _{ 5 }{ x }  }=64$ is,

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

  1. $5$

  2. $25$

  3. $125$

  4. $625$

Show answer
Correct Option: D
Explanation:

$3{ x }^{ \log _{ 5 }{ 2 }  }+{ 2 }^{ \log _{ 5 }{ x }  }=64$


Since, ${ a }^{ \log _{ b }{ c }  }={ c }^{ \log _{ b }{ a }  }$

Therefore, $3\left( { 2 }^{ \log _{ 5 }{ x }  } \right) +{ 2 }^{ \log _{ 5 }{ x }  }=64$

$\Rightarrow 4 \cdot 2^{\log _5x}=64$

$\Rightarrow { 2 }^{ \log _{ 5 }{ x }  }=16={ 2 }^{ 4 }$

Equating power of 2

$\Rightarrow \log _{ 5 }{ x } =4$

$\Rightarrow x={ 5 }^{ 4 }=625$

Ans: D

Sometimes to solve an equation, we may use the identity ${ a }^{ \log _{ a }{ b }  }=b,b>0,a>0,a\neq 1$
Then the number of solution(s) of ${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$ is/are,

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

  1. $0$

  2. $1$

  3. $2$

  4. infinite

  5. None of these

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Correct Option: A
Explanation:

${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$


Above equation is valid when $x>0,x\neq 1$

${ x }^{ \log _{ x }{ { \left( x+3 \right)  }^{ 2 } }  }=16$
     
$\Rightarrow (x+3)^2=16=4^2, [\because { a }^{ \log _{ a }{ b }  }=b]$

$\Rightarrow x+3=\pm 4$

$\Rightarrow x=1,-7$

Since, $x>0,x\neq 1$

Therefore, $x=\left{ \emptyset  \right} $

Using the identity $\displaystyle a^{\log _{a}{n}}= n,$ find:
$\displaystyle 3^{-\tfrac{1}{2}\log _{3}9}$, 

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

  1. $0.33$

  2. $-0.33$

  3. $0.66$

  4. $-0.66$

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Correct Option: A
Explanation:
Given $a^{log _a n}=n$    ....(1)

Consider, $\displaystyle 3^{-\frac{1}{2}\log _{3}9}$
$= 3^{\log _{3}9^{-1/2}}$
$= 9^{-1/2}$        (by (1))
$= \displaystyle \frac{1}{3}$
$=0.33$

Ans: 0.33

Using the identity $\displaystyle a^{\log _{a}{n}}= n,$ find:

$\displaystyle 2^{2-\log _{2}5}$

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

  1. $0.6$

  2. $0.8$

  3. $0.2$

  4. $0.1$

Show answer
Correct Option: B
Explanation:
Given $\displaystyle a^{\log _{a}{n}}= n$      .....(1)

Consider, 

$\displaystyle 2^{2-\log _2 5}=$$\displaystyle 2^{2}.2^{-\log _{2}5}$

                    $= 4.2^{\log _{2}5^{-1}}$

                    $= 4\,.5^{-1}$       $(by (1))$

                    $= \cfrac{4}{5}$

$\displaystyle 2^{2-\log _2 5}=0.8$

Ans: 0.8

If $\displaystyle a^{\log _{a}10}= 10$, then the set of value(s) of $a$ is/are

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

  1. $\displaystyle a \in \left ( 0,1 \right )\cup \left ( 1,\infty \right )$

  2. $\displaystyle a \in \left [ 0,1 \right )\cup \left (1,\infty \right).$

  3. $\displaystyle a \in \left ( -1,0 \right )\cup \left ( 1,\infty \right )$

  4. $\displaystyle a \in \left (-1,0\right ]\cup \left ( 1,\infty \right ).$

Show answer
Correct Option: A
Explanation:

$\displaystyle a^{\log _{a}10}= 10$
$\displaystyle \log _a10.\log _{10}a = 1\Rightarrow \cfrac{\log _{10}a}{\log _{10}a}=1$
Given expression is correct for all $'a'$ provided $'a'$ belongs to the domain of the given expression.
$\displaystyle \log _{a}10$ is defined for a+ive, and $\displaystyle a\neq 1.$
$\displaystyle \therefore a \in \left ( 0,1 \right )\cup \left ( 1,\infty  \right)$

If $\displaystyle \log _{p}q+\log _{q}r+\log _{r}p$ vanishes, where $p,q$ and $r$ are positive reals different than unity, then the value of $\displaystyle \left ( \log _{p}q \right )^{3}+\left ( \log _{q}r \right )^{3}+\left ( \log _{r}p \right )^{3} $ is

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

  1. an odd prime.

  2. an even number.

  3. an odd composite.

  4. an irrational number.

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Correct Option: A
Explanation:

Given 
$\displaystyle \log _{p}q+\log _{q}r+\log _{r}p=0$
$\Rightarrow \log _{p}q+\log _{q}r=-\log _{r}p$    ....(1)

Consider,$\left( \log _{ p } q \right) ^{ 3 }+\left( \log _{ q } r \right) ^{ 3 }+\left( \log _{ r } p \right) ^{ 3 }$

$=[(\log _{ p } q+\log _{ q } r)^{ 3 }-3\log _{ p } q\log _{ q } r(\log _{ p } q+\log _{ q } r)]+\left( \log _{ r } p \right) ^{ 3 }$

$=\left( -\log _{ r } p \right) ^{ 3 }-3\log _{ p } q\log _{ q } r\left( -\log _{ r } p \right) +\left( \log _{ r } p \right) ^{ 3 }$       (by (1))

$=3\log _{ p } q\log _{ q } r\log _{ r } p = 3 \quad \quad [\because \log _ab=\dfrac{\log b}{\log a}]$

The value of ${\left(\displaystyle\frac{1}{2}\right)}^{\log _{2}5}$ is equal to

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

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

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

  3. $ \displaystyle\frac{-1}{25}$

  4. $ \displaystyle\frac{1}{25}$

Show answer
Correct Option: A
Explanation:

$\displaystyle { \frac { 1 }{ 2 }  }^{ \log _{ 2 }{ 5 }  }={ 2 }^{ -\log _{ 2 }{ 5 }  }={ 2 }^{ \log _{ 2 }{ { 5 }^{ -1 } }  }={ 5 }^{ -1 }[\because a^{\log _ab}=b]=\frac { 1 }{ 5 } $

The value of $\displaystyle 49^{A}+5^{B}$, where $\displaystyle A= 1-\log _{7}2$ and $\displaystyle B= -\log _{5}4$ is

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

  1. $\displaystyle \frac{25}{2}$

  2. $\displaystyle \frac{49}{4}$

  3. $12$

  4. none of these

Show answer
Correct Option: A
Explanation:

$\displaystyle 49^{A}= 49^{1-\log _{7}2}= \frac{49}{49^{\log _{7}2}}= \frac{49}{7^{\log _{7}4}}= \frac{49}{4}\quad [\because a^{\log _ab}=b]$
$\displaystyle 5^{B}= 5^{-\log _{5}4} \quad [\because m\log a=\log a^m]=5^{\log _5\dfrac{1}{4}}= \frac{1}{4}$
$\therefore 49^A+5^B = \cfrac{25}{2}$

The value of the expression
$\displaystyle\frac{1}{1+\log _b\,a+\log _b\,c}+\displaystyle\frac{1}{1+\log _c\,a+\log _c\,b}+\displaystyle\frac{1}{1+\log _a\,b+\log _a\,c}$ is equal to

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

  1. $\,\,abc$

  2. $\,\,\displaystyle\frac{1}{abc}$

  3. $\,\,0$

  4. $\,\,1$

Show answer
Correct Option: D
Explanation:

$\displaystyle\frac{1}{1+\log _b\,a+\log _b\,c}+\displaystyle\frac{1}{1+\log _c\,a+\log _c\,b}+\displaystyle\frac{1}{1+\log _a\,b+\log _a\,c}$
$=\displaystyle \dfrac{1}{\displaystyle 1+\frac{\log a}{\log b}+\frac{\log c}{\log b}}+\dfrac{1}{\displaystyle 1+\frac{\log a}{\log c}+\frac{\log b}{\log c}}+\dfrac{1}{\displaystyle 1+\frac{\log b}{\log a}+\frac{\log c}{\log a}} [\because \log _yx=\dfrac{\log x}{\log y}]$
$=\displaystyle\frac{\log\,b}{\log\,b+\log\,a+\log\,c}+\displaystyle\frac{\log\,c}{\log\,c+\log\,a+\log\,b}+\displaystyle\frac{\log\,a}{\log\,c+\log\,a+\log\,b}\,=\,1$

Ans: D