Tag: maths

Questions Related to maths

The value of $\left (\dfrac {a^{-2} \times b^{-3}}{a^{-3}\times b^{-4}}\right )$ is _________.

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

  1. $a^{-1}\times b$

  2. $a \times b^{-1}$

  3. $(ab)^{-1}$

  4. $ab$

Show answer
Correct Option: D
Explanation:
We need to find value of $\left (\dfrac {a^{-2} \times b^{-3}}{a^{-3}\times b^{-4}}\right )$
By using $\dfrac {a^m}{a^n}=a^{m-n}$
Then it can be written as,
$a^{-2-(-3)}\times b^{-3-(-4)}$ $=$ $ab$   
Hence, option D is correct.

If $(\sqrt{2})^x + (\sqrt{3})^x = (\sqrt{13})^{\frac{x}{2}}$, then the value of $x$ is ___.

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

  1. $1$

  2. $2$

  3. $4$

  4. $0$

Show answer
Correct Option: C
Explanation:

$(\sqrt2)^x+(\sqrt3)^x=(\sqrt{13})^{\frac{x}{2}}$

$\Rightarrow 2^{\frac{x}{2}}+3^{\frac{x}{2}}=13^{\frac{x}{4}}$

$x$ should be the multiple of $4$.
If we put $x=4$
L.H.S$: 2^2+3^2=13$
and R.H.S $: 13^{\frac{4}{4}}=13$
$\therefore x=4$

If $a^2bc^3=5^3$ and $ab^2=5^6$, then $abc$ equals ___.

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

  1. $5$

  2. $5^2$

  3. $5^3$

  4. $5^{4.5}$

Show answer
Correct Option: C
Explanation:

$a^2bc^3=5^3$   ......(1)

$ab^2=5^6$      .......(2)
Multiplying equation 1 and 2, we get
$a^3b^3c^3=5^9$
$abc=5^{9/3}=5^3$

The value of $x$, if $5^{x-3}.3^{2x-a} = 225$ is ____.

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

  1. $3$

  2. $4$

  3. $2$

  4. $5$

Show answer
Correct Option: D
Explanation:

$225=3^2.5^2=5^{x-3}.3^{2x-a}$

Comparing powers of 5 on both sides 
$x-3=2$
$x=5$

The rationalising factor of $\sqrt[5]{a^2b^3c^4}$ is _____.

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

  1. $\sqrt[5]{a^3b^2c}$

  2. $\sqrt[5]{a^3bc}$

  3. $\sqrt[5]{a^3b^2c^5}$

  4. $\sqrt[5]{a^3b^6c}$

Show answer
Correct Option: A
Explanation:

To rationalize $(a^2b^3c^4)^{\frac{1}{5}}$, fifth root must be removed, 

$\therefore$We should multiply it by the factor $(a^3b^2c)^{\frac{1}{5}}$, So thst it will become $abc$.

$\left(\dfrac{5^a}{5^b}\right)^{a+b}.\left(\dfrac{5^b}{5^c}\right)^{b+c}.\left(\dfrac{5^c}{5^a}\right)^{c+a} =$ 

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

  1. $1$

  2. $4$

  3. $5$

  4. $0$

Show answer
Correct Option: A
Explanation:

We have, $\Bigr(\dfrac{5^a}{5^b}\Bigl)^{a+b}\cdot\Bigl(\dfrac{5^b}{5^c} \Bigr)^{b+c}\cdot \Bigl(\dfrac{5^c}{5^a} \Bigr)^{c+a}$


$=(5^{a-b})^{a+b}\cdot(5^{b-c})^{b+c}\cdot(5^{c-a})^{c+a}$

$=5^{a^2-b^2}\cdot 5^{b^2-c^2}\cdot 5^{c^2-a^2}\ $

$=\dfrac{5^{a^2}}{5^{b^2}}\cdot \dfrac{5^{b^2}}{5^{c^2}}\cdot \dfrac{5^{c^2}}{5^{a^2}}\\=1$

Comparing the numbers $10^{-49}$ and 2. $10^{-50}$ we may say

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

  1. the first exceeds the second by 8. $10^{-1}$

  2. the first exceeds the second by 2. $10^{-1}$

  3. the first exceeds the second by 8. $10^{-50}$

  4. the second is five times the first

  5. the first exceeds the second by 5

Show answer
Correct Option: C
Explanation:

${ 10 }^{ -49 }-2\cdot { 10 }^{ -50 }={ 10 }^{ -50 }(10-2)=8\cdot { 10 }^{ -50 }\ \therefore { 10 }^{ -49 }\hspace{1mm} exceeds\hspace{1mm} 2\cdot { 10 }^{ -50 }\hspace{1mm} by\hspace{1mm} 8\cdot { 10 }^{ -50 }$

If ${2^a} = 3$ and ${9^b} = 4$ then the value of $a.b$ is

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:
 ${ 2 }^{ a }=3$$ a\log _{ 10 }{ 2 } =\log _{ 10 }{ 3 } $$ a= \dfrac { \log _{ 10 }{ 3 }  }{ \log _{ 10 }{ 2 }  }   $  ${ 9 }^{ b }=4$$ b\log _{ 10 }{ 9 } =\log _{ 10 }{ 4 } $$b=\dfrac { \log _{ 10 }{ 4 }  }{ \log _{ 10 }{ 9 }  } $$ b=\dfrac { \log _{ 10 }{ { 2 }^{ 2 } }  }{ \log _{ 10 }{ { 3 }^{ 3 } }  } $$ b=\dfrac { \log _{ 10 }{ { 2 } }  }{ \log _{ 10 }{ { 3 } }  } $

$\therefore a.b= \dfrac { \log _{ 10 }{ 3 }  }{ \log _{ 10 }{ 2 }  }   \times \dfrac { \log _{ 10 }{ { 2 } }  }{ \log _{ 10 }{ { 3 } }  }$


$\therefore a.b=1$

whether the following relation is${{ \frac{1}{{{x^{a - b}}}}} ^{\frac{1}{{a - c}}}}{{ \frac{1}{{{x^{b - c}}}}} ^{\frac{1}{{b - a}}}}{{ \frac{1}{{{x^{c - a}}}}} ^{^{\frac{1}{{c - b}}}}} = 1$

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

  1. True

  2. False

Show answer
Correct Option: A

If ${2^{n-m}}=16$ and $3^{n+m}=729$  then $mn=?$

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

  1. $5$

  2. $4$

  3. $6$

  4. None of the above.

Show answer
Correct Option: A
Explanation:

$2^{n-m}=16\2^{n-m}=2^4\n-m=4\cdots(1)\3^{n+m}=729\3^{n+m}=3^6\n+m=6\cdots(2)\(1)+(2)\2n=10\\boxed{n=5}\\boxed{m=1}$
$\boxed{nm=5}...Answer$