Tag: a relation between logarithmic functions

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Questions Related to a relation between logarithmic functions

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

The value of $\,3^{\textstyle \log _4\,5}\,+\,4^{\textstyle \log _5\,3}\,-5^{\textstyle \log _4\,3}\,-3^{\textstyle \log _5\,4}$ is equal to

  1. $\,\,0$
  2. $\,\,1$
  3. $\,\,2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Use $\,a^{\textstyle \log _b\,c}\,=\,c^{\textstyle \log _b\,a}$
So, $[5^{\textstyle \log _4\,3}=3^{\textstyle \log _4\,5}]$
and $[4^{\textstyle \log _5\,3}=3^{\textstyle \log _5\,4}]$
$\Rightarrow\,\,3^{\textstyle \log _4\,5}\,+\,4^{\textstyle \log _5\,3}\,-\,3^{\textstyle \log _4\,5}\,-\,4^{\textstyle \log _5\,3}\,=\,0$

Ans: A
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

If $\left( \log _{ 3 }{ x }  \right) \left( \log _{ x }{ 2x }  \right) \left( \log _{ 2x }{ y }  \right) =\log _{ x }{ { x }^{ 2 } } $, then what is $y$ equal to?

  1. $4.5$
  2. $9$
  3. $18$
  4. $27$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$(\log _{ 3 }{ x }) (\log _{ x }{ 2x }) (\log _{ 2x }y)=\log _{x} x^{2}$ .... $(i)$
$(\log _{ 3 }{ x }) (\log _{ x }{ 2x }) (\log _{ 2x }y)$
$=\left(\dfrac { \log x}{\log 3}\right) \left(\dfrac {\log 2x}{\log x} \right) \left(\dfrac { \log y}{\log 2x}\right)$ .......[using base change formula]
$=\dfrac { \log { y }  }{ \log { 3 }  } =\log _{ 3 }{ y } $
Also, $\log _{ x }{ { x }^{ 2 } } =2\log _{ x }{ x } =2$
So, $\log _{ 3 }{ y } =2$ .... From $(i)$
$\Rightarrow y={ 3 }^{ 2 }=9$
Hence, option B is correct

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

If $\displaystyle \log _{ 2a }{ a } =x$, $\log _{ 3a }{  2a } =y$ and $\log _{ 4a }{  3a } =z$, then $xyz-2yz$ is equal to

  1. 1

  2. -1

  3. 0

  4. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

${{\log } _{2a}}a=x,{{\log } _{3a}}2a=y,{{\log } _{4a}}3a=z$

$ {{\log } _{2a}}a=x $

$ \dfrac{\log a}{\log 2a}=x $

Similarly,

$ {{\log } _{3a}}2a=y $

$ \dfrac{\log 2a}{\log 3a}=y $

Similarly,

$ {{\log } _{4a}}3a=z $

$ \dfrac{\log 3a}{\log 4a}=z $


Therefore,,

$ =xyz-2yz $

$ =\dfrac{\log a}{\log 2a}\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log 3a}{\log 4a}-2\times \dfrac{\log 2a}{\log 3a}\times \dfrac{\log 3a}{\log 4a} $

$ =\dfrac{\log a}{\log 4a}-2\times \dfrac{\log 2a}{\log 4a} $

$ =\log a-\log 4-\log a-2\times \left( \log 2a-\log 4a \right) $

 $ =-\log 4-2\times \left( \log 2a-\log 2a-\log 2 \right) $

 $ =-\log 4+2\log 2 $

 $ =-\log 4+\log 4 $

 $ =0 $

So, the value is 0.