Tag: laws of exponents and powers

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Questions Related to laws of exponents and powers

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

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

  1. $a^{-1}\times b$
  2. $a \times b^{-1}$
  3. $(ab)^{-1}$
  4. $ab$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

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

  1. $1$
  2. $2$
  3. $4$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $: 13^{\frac{4}{4}}=13$
$\therefore x=4$

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

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

  1. $\sqrt[5]{a^3b^2c}$
  2. $\sqrt[5]{a^3bc}$
  3. $\sqrt[5]{a^3b^2c^5}$
  4. $\sqrt[5]{a^3b^6c}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $abc$.

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

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

  1. $1$
  2. $4$
  3. $5$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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}\ $

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

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 } }  } $

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


$\therefore a.b=1$

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

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

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using exponent rules, each term simplifies to x^(b-a)/(a-c) * x^(c-b)/(b-a) * x^(a-c)/(c-b). Adding the exponents (b-a)/(a-c) + (c-b)/(b-a) + (a-c)/(c-b) results in 0, and x^0 = 1.