Tag: maths

Questions Related to maths

The value of $\log _{ 3 }{ 9 } +\log _{ 5 }{ 25 } +\log _{ 2 }{ 8 } $ is

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

  1. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: D
Explanation:

$\log _{ 3 }{ 9 } +\log _{ 5 }{ 25 } +\log _{ 2 }{ 8 } =\log _{ 3 }{ { \left( 3 \right)  }^{ 2 } } +\log _{ 5 }{ { \left( 5 \right)  }^{ 2 } } +\log _{ 2 }{ { \left( 2 \right)  }^{ 3 } } =2\log _{ 3 }{ { \left( 3 \right)  }^{  } } +2\log _{ 5 }{ { \left( 5 \right)  }^{  } } +3\log _{ 2 }{ { \left( 2 \right)  }^{  } } =7$

Solve the following: $\dfrac{1}{\log _{xy} \, xyz} \, + \, \dfrac{1}{\log _{xz} \, xyz} \, + \, \dfrac{1}{\log _{zx} \, xyz} \, =$

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

  1. 0

  2. 1

  3. 2

  4. $log _x \, xyz$

Show answer
Correct Option: C
Explanation:

We have,

$\dfrac{1}{\log _{xy} \, xyz} \, + \, \dfrac{1}{\log _{yz} \, xyz} \, + \, \dfrac{1}{\log _{zx} \, xyz}$

$\Rightarrow \dfrac{\log _{xy}xy}{\log _{xy} \, xyz} \, + \, \dfrac{\log _{yz}yz}{\log _{yz} \, xyz} \, + \, \dfrac{\log _{zx}zx}{\log _{zx} \, xyz} $

We know that
$\dfrac{\log _a m}{\log _a n}=\log _n m$

Therefore,
$\Rightarrow \log _{xyz}xy\, + \, \log _{xyz}yz\, + \,\log _{xyz}zx$

$\Rightarrow \log _{xyz}(x^2y^2z^2)$

$\Rightarrow \log _{xyz}(xyz)^2$

$\Rightarrow 2\log _{xyz}(xyz)$

$\Rightarrow 2$

Hence, this is the answer.

If $(150)^x = 7$, then x is equal to:

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

  1. $\displaystyle \frac{log 7}{(log 3)+(log 5)+1}$

  2. $\displaystyle \frac{log7}{(log3)+(log6)}$

  3. $\displaystyle \frac{log7}{(log3)+(log5)+10}$

  4. $\displaystyle \frac{log7}{log2+log3}$

Show answer
Correct Option: A
Explanation:

$(150)^x = 7$


$\Rightarrow log _{150} 7 = x$

$\Rightarrow \dfrac{log 7}{log 150} = x$

$\Rightarrow \dfrac{log 7}{log(3\times 5\times 10)} = \dfrac{log 7}{log 3+log 5+ log 10}$

$= \dfrac{log 7}{log 3+log 5+1}$

Given that $N = 7^{\log _{49} 900} , A = 2^{\log _{2} 4} + 3^{\log _{2} 4} + 4^{\log _{2} 2} - 4^{\log _{2} 3} , D = (\log _5\, 49) (\log _7 \, 125)$ 
Then answer the following questions : (using the values of $N, A, D$)
If $\log _A \, D = a$, then the value of $\log _6 \, 12$ is (in terms of $a$)

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

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

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

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

  4. $\dfrac{1 + 3a}{2a}$

Show answer
Correct Option: A
Explanation:
$A=2^{\log _{2}4}+3^{\log _{2}4}+4^{\log _{2}2}-4^{\log _{2}3}$
$A=2^{\log _{2}(2)^{2}}+3^{\log _{2}(2)^{2}}+4^{\log _{2}2}-3^{\log _{2}4}$             ($a^{\log _{x}b}=b^{\log _{x}a}$)
$A=2^{2\log _{2}2}+3^{2\log _{2}2}+4^1-3^{2\log _{2}2}$
$A=2^{2}+3^{2}+4-3^{2}=8$

$D=(\log _{5}49)(\log _{7}125)$
$D=(\log _{5}(7)^{2})(\log _{7}(5)^{3})$
$D=2\log _{5}7\times 3\log _{7}5$
$D=2\times 3=6$                    (  $\log _{5}7=\dfrac{1}{\log _{7}5}$  )

It is given that,
$\log _{A}D=a$
$\log _{8}6=a$
$\log _{6}8=\dfrac{1}{a}$
$\log _{6}(2)^{3}=\dfrac{1}{a}$
$\log _{6}2=\dfrac{1}{3a}$

Now,
$\log _{6}12=\log _{6}\left ( 6\times 2 \right )$
$=\log _{6}6+\log _{6}2$

$=1+\dfrac{1}{3a}$
$=\dfrac{3a+1}{3a}$







If $A = \log _2 \, \log _2 \, \log _4 \, 256 + 2 \, \log _{\sqrt{2}} \, 2$, then $A$ is equal to 

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

  1. $2$

  2. $3$

  3. $5$

  4. $7$

Show answer
Correct Option: C
Explanation:
Use the properties
$\log _{ a }{ { \left( m \right)  }^{ n }=n\log _{ a }{ m }  } $  and  $\log _{ a }{ a } =1$

$\log _{ 2 }{ \log _{ 2 }{ \log _{ 4 }{ 256 }  }  } +2\log _{ \sqrt { 2 }  }{ 2 } =\log _{ 2 }{ \log _{ 2 }{ \log _{ 4 }{ { \left( 4 \right)  }^{ 4 } }  }  } +2\log _{ \sqrt { 2 }  }{ { \left( \sqrt { 2 }  \right)  }^{ 2 } } $

$=\log _{ 2 }{ \log _{ 2 }{ 4 }  } +2\left( 2 \right) $
$=\log _{ 2 }{ \log _{ 2 }{ { \left( 2 \right)  }^{ 2 } }  } +4$
$=\log _{ 2 }{ 2 } +4$
$=1+4=5$

If $3{x^{{{\log } _5}2}} + {2^{{{\log } _5}x}} =64$ then $x$ is equal to

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

  1. $625$ 

  2. $250$

  3. $125$

  4. None of these

Show answer
Correct Option: A
Explanation:
$3{x}^{\log _{5}{2}}+{2}^{\log _{5}{x}}=64$

$\Rightarrow\,3 \times {2}^{\log _{5}{x}}+{2}^{\log _{5}{x}}=64$ using 

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

$\Rightarrow\,4\times {2}^{\log _{5}{x}}=64$

$\Rightarrow\,{2}^{2}\times {2}^{\log _{5}{x}}={2}^{6}$

$\Rightarrow\,2+\log _{5}{x}=6$  .................... taking log on both sides 

$\Rightarrow\,\log _{5}{x}=6-2=4$

$\Rightarrow\,x=5^4=625$

$\log _a {bc}= x, \log _b {ac}= y , \log _c {ab}= z$, 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,
${ 3 }^{ \log _{ 4 }{ 5 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  }$
${ =5 }^{ \log _{ 4 }{ 3 }  }-{ 5 }^{ \log _{ 4 }{ 3 }  },$ ($\because { x }^{ \log _{ a }{ y }  }={ y }^{ \log _{ a }{ x }  }$)
$=0$