Tag: indices

Questions Related to indices

The value of ${ \left[ { \left( { 3 }^{ 2 } \right)  }^{ 2 } \right]  }^{ -1 }$ is-

maths indices negative indices law of indices laws of indices

  1. $81$

  2. $-81$

  3. $-0.0123$

  4. $0.0123$

Show answer
Correct Option: D
Explanation:
$\cfrac { 1 }{ { \left( { 3 }^{ 2 } \right)  }^{ 2 } } $        $\because (a^m)^n=a^{mn}$
$=\cfrac { 1 }{ { 3 }^{ 4 } }$
$ =\cfrac { 1 }{ 81 }$
$ =0.0123$

The approximate value of ${ \left{ { { \left( 3.92 \right)  }^{ 2 }\quad +3\left( 2.1 \right)  }^{ 4 } \right}  }^{ { 1 }/{ 6 } }$

maths indices exponentiation index notation and products of prime factors powers

  1. $2.0466$

  2. $2.755$

  3. $2.345$

  4. $0.242718$

Show answer
Correct Option: A
Explanation:

${(3.92^2+3(2.1^4)}^\dfrac{1}{6}$

 
$=(15.3664+3\times19.4481)^\dfrac{1}{6}$


$=(15.3664+58.3443)^\dfrac{1}{6}$

$=(73.7107)^\dfrac{1}{6}$

$=2.0476$

If ${2}^{1998}-{2}^{1997}-{2}^{1996}+{2}^{1995}={K.2}^{1995}$, then the value of $K$ is 

maths indices exponentiation index notation and products of prime factors powers

  1. 3

  2. 2

  3. -2

  4. -3

Show answer
Correct Option: A
Explanation:
$2^{1998}-2^{1997}-2^{1996}+2^{1995}=k. 2^{1995}$
$2^{1995}(2^3-2^2-2^1+1)=k. 2^{1995}$
$(2^3-2^2-2^1+1)=k$
$(8-4-2+1)=k$
$k=3$

If $\displaystyle { 2 }^{ n }-{ 2 }^{ n-1 }=4$, then the value of $\displaystyle { n }^{ n }$ will be -

maths indices exponentiation index notation and products of prime factors powers

  1. 1

  2. $\displaystyle \frac { 3 }{ 2 } $

  3. 2

  4. 27

Show answer
Correct Option: D
Explanation:

$\displaystyle { 2 }^{ n }-{ 2 }^{ n-1 }=4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }\left( 2-1 \right) =4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }={ 2 }^{ 2 }$
$\displaystyle \therefore \quad n-1=2$
$\displaystyle \therefore \quad n=3$
$\displaystyle \therefore \quad { n }^{ n }={ 3 }^{ 3 }=27$

$\displaystyle \frac { { \left( 3.63 \right)  }^{ 2 }-{ \left( 2.37 \right)  }^{ 2 } }{ 3.63+2.37 } $ is simplified to -

maths indices exponentiation index notation and products of prime factors powers

  1. 6

  2. 1.36

  3. 2.26

  4. 1.26

Show answer
Correct Option: D
Explanation:

$\displaystyle \frac { { \left( 3.63 \right)  }^{ 2 }-{ \left( 2.37 \right)  }^{ 2 } }{ 3.63+2.37 } $
$\displaystyle =\frac { \left( 3.63+2.37 \right) \left( 3.63-2.37 \right)  }{ 3.63+2.37 } $
$\displaystyle =3.63-2.37$
$\displaystyle =1.26$

Value of $\displaystyle\frac{2^{100}}{2}$ is

maths indices exponentiation index notation and products of prime factors powers

  1. $1$

  2. $\displaystyle 50^{100}$

  3. $\displaystyle 2^{50}$

  4. $\displaystyle 2^{99}$

Show answer
Correct Option: D
Explanation:

$\displaystyle 2^{100}\div 2^1=2^{100-1}=2^{99}$

Simplest form of the Expression $\displaystyle { \left( { x }^{ 6 }.{ y }^{ { -5 }/{ 4 } } \right)  }^{ { -4 }/{ 3 } }$ will be-

maths indices exponentiation index notation and products of prime factors powers

  1. $\displaystyle { x }^{ -24 }y$

  2. $\displaystyle { x }^{ -8 }{ y }^{ { 5 }/{ 3 } }$

  3. $\displaystyle { x }^{ 8 }{ y }^{ { -5 }/{ 3 } }$

  4. $\displaystyle { x }^{ -8 }{ y }^{ { -5 }/{ 3 } }$

Show answer
Correct Option: B
Explanation:

$\displaystyle { \left( { x }^{ 6 }.{ y }^{ { -5 }/{ 4 } } \right)  }^{ { -4 }/{ 3 } }$
$\displaystyle ={ x }^{ -6\times { 4 }/{ 3 } }{ y }^{ \dfrac { -5 }{ 4 } \times -\dfrac { 4 }{ 3 }  }$
$\displaystyle ={ x }^{ -8 }{ y }^{ { 5 }/{ 3 } }$