Tag: maths

Questions Related to maths

Simplify and give reasons:
${ \left( \cfrac { 1 }{ 2 }  \right)  }^{ -3 }\times { \left( \cfrac { 1 }{ 4 }  \right)  }^{ -3 }\times { \left( \cfrac { 1 }{ 5 }  \right)  }^{ -3 }\quad $

maths power and exponent power of powers laws of exponents and powers law of indices

  1. ${40}^{3}$

  2. ${40}^{-3}$

  3. ${40}^{6}$

  4. None of these

Show answer
Correct Option: A
Explanation:

we know,

$(\dfrac{a}{b})^{-m}=(\dfrac{b}{a})^{m}$
So,
$(\dfrac{1}{2})^{-3}=(\dfrac{2}{1})^{3}=2^{3}$

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

$(\dfrac{5}{1})^{-3}=(\dfrac{5}{1})^{3}=5^{3}$
now we know,

\$a^{m}b^{m}*c^{m}=(abc)^{m}$

$\implies(2)^{3}(4)^{3}*(5)^{3}$

$=(40)^{3}$

$(-2)^{-5}\times (-2)^{6}$ is equal to

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $-2$

  2. $2$

  3. $-5$

  4. $6$

Show answer
Correct Option: A
Explanation:

As we know that $a^{m}\times a^{n}=a^{(m+n)}$


 So, $(-2)^{-5}\times (-2)+^{6}=(-2)^{-5+6}$ 

        $(-2)^{-5}\times (-2)+^{6}=(-2)^{1}$ 

        $(-2)^{-5}\times (-2)+^{6}=-2$

$(-1)^{50}$ is equal to

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $-1$

  2. $50$

  3. $-50$

  4. $1$

Show answer
Correct Option: D
Explanation:

As we can see that the power is even and we know that when $(-1)$ is raised to some odd power then its value remains same i.e. $-1$ but when the power is even its sign is changed i.e. $1$ . therefore, $(-1)^{50}=1$

$(-2)^{-2}$ is equal to

maths power and exponent power of powers laws of exponents and powers law of indices

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

  2. $\dfrac {1}{4}$

  3. $\dfrac {-1}{2}$

  4. $\dfrac {-1}{4}$

Show answer
Correct Option: B
Explanation:
 As we know that $a^{-b}$ is equal to $1/a^{b}.$ So, $(-2)^{-2}$ is equal to $1/(-2)^{2}$.
 
 Also,we know that $(-2)^{2}=(-2)\times (-2)=4$ 

 Therefore , $(-2)^{-2}=1/4$

Choose the correct option:
$\left[\dfrac{{100}}{{101}}\right]^3$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $\dfrac{{100}^3}{{101}^3}$

  2. $\dfrac{{100}^4}{{101}^4}$

  3. $\dfrac{{1000}^2}{{101}^2}$

  4. $\dfrac{{100}}{{101}}$

Show answer
Correct Option: A
Explanation:

$Now\quad \left[ \dfrac { 100 }{ 101 }  \right] ^{ 3 }\quad \ \quad \quad =\quad \dfrac { { 10 }0^{ 3 } }{ 101^{ 3 } } \quad \left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) \ $

Choose the correct options:$\dfrac{{10}^2}{{11}^2}$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $\left[\dfrac{{10}}{{11}}\right]^2$

  2. $\left[\dfrac{{100}}{{11}}\right]^2$

  3. $\left[\dfrac{{10}}{{11}}\right]^4$

  4. $\left[\dfrac{{5}}{{11}}\right]^2$

Show answer
Correct Option: A
Explanation:

$Now\quad \dfrac { { 10 }^{ 2 } }{ 11^{ 2 } } \ =\quad \left( \dfrac { 10 }{ 11 }  \right) ^{ 2 }\left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) $

Choose the correct option:
$\left(\dfrac{5^5\times6^5}{3^5}\right)$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $\left(\dfrac{5\times6}{3}\right)^5$

  2. $\left(\dfrac{5\times6}{3}\right)^6$

  3. $\left(\dfrac{5\times6}{5}\right)^3$

  4. $\left(\dfrac{5\times6}{5}\right)^5$

Show answer
Correct Option: A
Explanation:

$Now\quad \left( \dfrac { { 5 }^{ 5 }\times { 6 }^{ 5 } }{ { 3 }^{ 5 } }  \right) \quad \ \quad \quad =\quad \left( \dfrac { 5\times 6 }{ 3 }  \right) ^{ 5 }\quad \left( \because \left( \dfrac { { a }^{ m }\times { c }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a\times c }{ b }  \right) ^{ m } \right) $

The value of $(15)^4$ is equal to:
maths power and exponent power of powers laws of exponents and powers law of indices

  1. $(15)^4=3^2.5^2$

  2. $(15)^4=3^4.5^4$

  3. $(15)^4=3^3.5^3$

  4. $(15)^4=3^2.5^3$

Show answer
Correct Option: B
Explanation:
We need to find value of $(15)^4$
It can be written as $(15)^4=(3\times 5)^4$
$=3^4\times 5^4$    ....Using law $(ab)^m=a^m.b^m$
Hence, option B is correct.

Simplify the following using law of exponents.
$\dfrac{9^7}{9^{15}}$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $9^{-8}$

  2. $\dfrac{1}{9^8}$

  3. $9^8$

  4. $9^{1/8}$

Show answer
Correct Option: A,B
Explanation:

we know,


$\dfrac{a^{m}}{a^{n}}=a^{m-n}$

so,

$\dfrac{9^{7}}{9^{15}}$

$=9^{7-15}$

$=9^{-8}$

$=\dfrac{1}{9^{8}}$

Simplify the following using law of exponents.
$(-6^4)^4$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $(-6)^{16}$

  2. $(-6)^0$

  3. $(-6)^8$

  4. $(-6)^1$

Show answer
Correct Option: A
Explanation:

we know,


$(a^{m})^{n}=a^{mn}$

so,

$((-6)^{4})^{4}=(-6)^{4*4}$

$=(-6)^{16}$