Tag: logarithms

Quizzes (3) Questions (108)

Questions Related to logarithms

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

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. $\dfrac{\ln a}{\ln m}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

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

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

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. $\displaystyle \frac{\log y}{\log \{ x(1 + a) \}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle p = \dfrac{\displaystyle log _e\left ( \frac{y}{x} \right )}{log _e (1 + a)}$

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

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

  1. an integer

  2. a prime number

  3. a rational number

  4. an irrational number
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

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. $y=6$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
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 ))} $ .......

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

$\therefore 2^y =32$

$y=5$