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Questions Related to maths

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 ))} $ .......

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

$\therefore 2^y =32$

$y=5$

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 $y$ equals:

  1. $\cfrac{9}{2}$
  2. $9$
  3. $18$
  4. $27$
  5. $81$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $\log _{ a }{ b } =\cfrac { \log _{ c }{ b }  }{ \log _{ c }{ a }  } $, we have with base $x$, 
$\cfrac { \log { x }  }{ \log { 3 }  } .\cfrac { \log { 2x }  }{ \log { x }  } .\cfrac { \log { y }  }{ \log { 2x }  } =2;\quad \quad \log { y } =2\log { 3 } =\log { 9 } ;\quad y=9$

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

If $\log _{2x}$$216= x$, where $x$ is real, then $x$ is:

  1. A non-square, non-cube integer

  2. A non-square, non-cube, non-integral number

  3. An irrational number

  4. A perfect square

  5. A perfect cube
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\log _{2x}216= x , (2x)^x = 216 , 2^x.x^x= 2^3 . 3^3$
An obvious solution to this eqquation is $x = 3$, so that (A) is the correct choice.