Tag: laws of exponents and powers

Quizzes (0) Questions (94)

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 ${({3}^{m})}^{n}$, for every pair of integers $(m,n)$ is 

  1. ${3}^{m+n}$
  2. ${3}^{mn}$
  3. ${3}^{{m}^{n}}$
  4. ${3}^{m}+{3}^{n}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that for some positive integers $m$ and $n$, 

$\left( x^{ m } \right) ^{ n }=x^{ m\times n }=x^{mn}$
Therefore, $\left( 3^{ m } \right) ^{ n }=3^{ m\times n }=3^{mn}$
Hence, the value of $\left( 3^{ m } \right) ^{ n }$ is $3^{mn}$.

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

$(2^{0} + 4^{-1})\times 2^{2}$ is equal to

  1. $2$
  2. $5$
  3. $4$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we know that $a^{-b}$ is equal to $1/a^{b}$. Also, $p^{0}=1$ 


So, $(2^{0}+4^{-1})\times 2^{2}=(1+1/4)\times 2^{2}$ 

by using distributive law of multiplication , we get

 $(2^{0}+4^{-1})\times 2^{2}=1\times 2^{2}+1/4\times 2^{2}$ 

$(2^{0}+4^{-1})\times 2^2=4+1/4\times 4$ (because $2^{2}=2\times 2=4$) 

$(2^{0}+4^{-1})\times 2^{2}=4+1=5$

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

The value of $\left(\dfrac{x^q}{x^r}\right)^{\dfrac{1}{qr}} \times \left(\dfrac{x^r}{x^p}\right)^{\dfrac{1}{rp}}\times \left(\dfrac{x^p}{x^q}\right)^{\dfrac{1}{pq}}$ is equal to ___.

  1. $x^{\frac{1}{p}+\frac{1}{q}+\frac{1}{2}}$
  2. $0$
  3. $x^{pq+qr+rp}$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(x^{q-r})^{\cfrac{1}{qr}}\times (x^{r-p})^{\cfrac{1}{rp}}\times (x^{p-q})^{\cfrac{1}{pq}} $


$=x^{\cfrac{q-r}{qr}}\times x^{\cfrac{r-p}{rp}}\times x^{\cfrac{p-q}{pq}}$
On adding all the powers of $x$, we get
$\Rightarrow x^{\bigl(\cfrac{q-r}{qr}+\cfrac{r-p}{rp}+\cfrac{p-q}{pq}\bigr)}$

$=x^{\cfrac{p(q-r)+q(r-p)+r(p-q)}{pqr}}$

$=x^{\cfrac{0}{pqr}}=1$

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

$\left(\dfrac{1}{x^{a-b}}\right)^{\tfrac{1}{(a-c)}}. \left(\dfrac{1}{x^{b-c}}\right)^{\tfrac{1}{(b-a)}}. \left(\dfrac{1}{x^{c-a}}\right)^{\tfrac{1}{(c-b)}}=$

  1. $0$
  2. $1$
  3. $a+b+c$
  4. $(a-b+c)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can write the given equation as, 

$(x^{b-a})^{\frac{1}{a-c}}\cdot (x^{c-b})^{\frac{1}{b-a}}\cdot (x^{a-c})^{\frac{1}{c-b}}$

$=x^{\cfrac{b-a}{a-c}}\cdot x^{\cfrac{c-b}{b-a}}\cdot x^{\cfrac{a-c}{c-b}}$
On adding all the powers of $x$, We get
$x^{\Bigl(\cfrac{(b-a)^2(c-b)+(c-b)^2(a-c)+(a-c)^2(b-a)}{(a-c)(b-c)(c-b)}\Bigr)}\ =x^0=1$