Tag: more about logarithms

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Questions Related to more about logarithms

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

Which is the correct order for a given number $\alpha$ in increasing order.

  1. $\log _{2} \alpha, \log _{e} \alpha, \log _{3} \alpha, \log _{10} \alpha$
  2. $\log _{10} \alpha, \log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha$
  3. $\log _{10} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{3} \alpha$
  4. $\log _{3} \alpha, \log _{e} \alpha, \log _{2} \alpha, \log _{10} \alpha$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that, greater the value of base lesser is the value of the logarithm

Also, $10<3<e(2.71)<2$

$\therefore $ For any positive number $\alpha $,

${ log } _{ 2 }\alpha >{ log } _{ e }\alpha >{ log } _{ 3 }\alpha >{ log } _{ 10 }\alpha $

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

If $\log _3{(\log _3{a})}+\log _{\cfrac{1}{3}}{\left(\log _{\cfrac{1}{3}}{b}\right)}=1$, then the value of $ab^3$ is 

  1. $9$
  2. $3$
  3. $1$
  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let x = log3(a) and y = log(1/3)(b). The equation is log3(x) + log(1/3)(y) = 1. Since log(1/3)(y) = -log3(y), we have log3(x) - log3(y) = 1, so log3(x/y) = 1, meaning x/y = 3, or x = 3y. Substituting back: log3(a) = 3 * log(1/3)(b) = 3 * (-log3(b)) = -3 * log3(b) = log3(b^-3). Thus, a = b^-3, which means a * b^3 = 1.

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 ))} $ .......

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

$\therefore 2^y =32$

$y=5$