Tag: law of indices

Questions Related to law of indices

$5^{-2}$ can also be expressed as

  1. $\dfrac{1}{25}$

  2. $\dfrac{1}{5^{-2}}$

  3. $\dfrac{1}{5}$

  4. None of these


Correct Option: A
Explanation:
$5^{-2}=5^{-1*2}$

$=(5^2)^{-1}$                       $\because  (a)^{mn}=(a^m)^n$

$=\dfrac{1}{5^2}$                              $\because a^{-n}=\dfrac{1}{a^n}$

$=\dfrac{1}{25}$

If the exponent of a negative integer is odd, then the result is a .......... integer.

  1. positive

  2. negative

  3. zero

  4. None of these


Correct Option: B
Explanation:

  If the exponent of a negative integer is odd, then the result is a negative integer.

$\Rightarrow$  $(-2)^{3}=(-2)\times (-2)\times (-2)=-8$
$\Rightarrow$  Here, $-2$ is negative base and $3$ is odd power.
$\Rightarrow$  Then result  $-8$ is a negative integer.

If the exponent of a negative integer is even then the result is a ............ integer.

  1. Positive

  2. Negative

  3. 0

  4. None


Correct Option: A
Explanation:

  If the exponent of a negative integer is even, then the result is a $positive$ integer.

$\Rightarrow$  $(-2)^{4}=(-2)\times (-2)\times (-2)\times(-2)=16$
$\Rightarrow$  Here, $-2$ is negative base and $4$ is even power.
$\Rightarrow$  Then result  $16$ is a positive integer.

Evaluate : $(-4)^{-2}$---

  1. $\displaystyle \frac{1}{-16}$

  2. $\displaystyle \frac{1}{16}$

  3. $-16$

  4. $16$


Correct Option: B
Explanation:
$(-4)^{-2}=\dfrac{1}{(-4)^2}$            $\because a^{-m}=\dfrac{1}{a^m}$

$=\dfrac{1}{-4*-4}$

$=\dfrac1{16}$

$\therefore(-4)^{-2}=\dfrac1{16}$

Which of the following has an exponent with negative index?

  1. $-3^{4}$

  2. $4^{3}$

  3. $\dfrac {1}{3^{4}}$

  4. $-3$


Correct Option: C
Explanation:

$\dfrac {1}{3^{4}} = 3^{-4}$

$\therefore \dfrac {1}{3^{4}}$ has a negative index.
So, option $C$ is correct.

Simplify: $1 \div 7 \div 7$

  1. $7^{-2}$

  2. $7^{-1}$

  3. $0.7$

  4. $0.777$


Correct Option: A
Explanation:

$1 \div 7 \div 7 = \dfrac 17 \div 7 = \dfrac 17 \times \dfrac 17$


$=\dfrac {1}{7^2} = 7^{-2}$

So, option $A$ is correct.

Which of the following has a negative index?

  1. ${ x }^{ 5 }$

  2. $\dfrac { 1 }{ x } $

  3. ${ (x) }^{ \tfrac { 3 }{ 2 } }$

  4. ${ (-x) }^{ \tfrac { 3 }{ 2 } }$


Correct Option: B
Explanation:

$\dfrac{1}{x} = x^{-1}$. Index of $x$ is $(-1)$, which is negative.


Hence, option $B$ is correct

Which of the following has a negative index?

  1. $ \dfrac { 1 }{ { x }^{ -3 } } $

  2. ${ x }^{ 3 }$

  3. ${ -x }^{ 3 }$

  4. $\dfrac { 1 }{ { x }^{ 3 } } $


Correct Option: D
Explanation:

$\dfrac{1}{x^3}=x^{-3}$ . Index of $x$ is  $(-3)$, which is negative.


Hence, option $D$ is correct.

Which of the following has a negative index?

  1. $\dfrac { 1 }{ { x }^{ 2 } } $

  2. ${ (x) }^{ \tfrac { 1 }{ 4 } }$

  3. ${ (-x) }^{ 2 }$

  4. $\dfrac { 1 }{ { x }^{ -2 } } $


Correct Option: A
Explanation:

$\dfrac { 1 }{ { x }^{ 2 } } $ = ${ x }^{ -2 }$
Thus ${ x }^{ -2 }$ has a negative index
Therefore option $A$ is the correct answer.

Which of the following has a negative index?

  1. ${ (x) }^{ \tfrac{ 1 }{4 } }$

  2. ${ (x) }^{ \tfrac { 2 }{ 3 } }$

  3. $\dfrac { 1 }{ { x }^{ 2 } } $

  4. ${ (-x) }^{\tfrac { 2 }{ 3 } }$


Correct Option: C
Explanation:
Option A has positive index i.e. $\dfrac {1}{4}$.
Option B also has positive index i.e., $\dfrac {2}{3}$
In option C, $\dfrac{1}{x^2} = x^{-2}$.  
Index of $x$ is $(-2)$, which is negative.
Hence, option C is correct