Tag: maths

Questions Related to maths

Given $\log _{ 10 }{ x } =a,\log _{ 10 }{ y } =b$

Write down ${ 10 }^{ a-1 }$ in terms of $x$.

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

  1. $\cfrac{10}{x}$

  2. $10x$

  3. $x$

  4. $\cfrac{x}{10}$

Show answer
Correct Option: D
Explanation:

$\log _{10} x = a$

$10^a = x$

$\therefore 10^{a – 1} = \dfrac{10^a}{10^ 1} = \dfrac{x}{10}$

Given $\log _{ 10 }{ x } =a,\log _{ 10 }{ y } =b$

Write down ${ 10 }^{ 2b }$

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

  1. ${y}^{a}$

  2. ${y}$

  3. ${y}^{2}$

  4. ${y}^{b}$

Show answer
Correct Option: C
Explanation:

$\log _{10} y = b$

$10^b = y$

 $ \therefore 10^{2b} = (10^b)^2 = y^2$

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

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

  1. ${3}^{m+n}$

  2. ${3}^{mn}$

  3. ${3}^{{m}^{n}}$

  4. ${3}^{m}+{3}^{n}$

Show answer
Correct Option: B
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 $3^{mn}$.

Simplify the following:
${(-5)}^{4}\times {(-5)}^{-6}$

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

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

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

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

  4. None of these

Show answer
Correct Option: A
Explanation:

we know that,


$\because a^m*a^n=a^{m+n}$

and

$\because \dfrac{1}{a^m}=a^{-m}$
${(-5)}^{4}\times {(-5)}^{-6}$

$=(-5)^{-6+4}$

$=(-5)^{-2}$

$=\dfrac{1}{(-5)^2}$

$=\dfrac1{25}$

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

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

  1. $2$

  2. $5$

  3. $4$

  4. $3$

Show answer
Correct Option: B
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 $2^{2}=2\times 2=4$) 

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

The value of $(4\times 5)^6$ is equal to:

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

  1. $4^6\times6^5$

  2. $4^6\times5^5$

  3. $4^6\times5^6$

  4. $4^6\times7^6$

Show answer
Correct Option: C
Explanation:
We need to find value of $(4\times 5)^6$ 

  $=4^6\times 5^6$      .....By using law $(a\times b)^m=a^m\times b^m$

Hence, option C is correct.

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 ___.

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

  1. $x^{\frac{1}{p}+\frac{1}{q}+\frac{1}{2}}$

  2. $0$

  3. $x^{pq+qr+rp}$

  4. $1$

Show answer
Correct Option: D
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
$\Rightarrow x^{\bigl(\cfrac{q-r}{qr}+\cfrac{r-p}{rp}+\cfrac{p-q}{pq}\bigr)}$

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

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

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

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

  1. $0$

  2. $1$

  3. $a+b+c$

  4. $(a-b+c)^2$

Show answer
Correct Option: B
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
$x^{\Bigl(\cfrac{(b-a)^2(c-b)+(c-b)^2(a-c)+(a-c)^2(b-a)}{(a-c)(b-c)(c-b)}\Bigr)}\ =x^0=1$

The $100^{th}$ root of $10^{(10^{10})}$ is ___.

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

  1. $10^{8^{10}}$

  2. $10^{10^{8}}$

  3. $(\sqrt{10})^{(\sqrt{10})^{10}}$

  4. $10(\sqrt{(10)})^{\sqrt{10}}$

Show answer
Correct Option: B
Explanation:

$100th $ root of $10^{(10^{10})}=(10^{(10^{10})})^{\frac{1}{100}}$


$\Rightarrow 10^{(\frac{10^{10}}{100})}=10^{(10^8)}$

Consider the following statements.
Assertion $(A): a^0 = 1, a\neq 0$
Reason $(R): a^m\div a^n = a^{m-n}$, where $m,n$ being integers.
Which of the following options hold?

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

  1. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

  2. Both $A$ and $R$ are true and $R$ is not the correct explanation of $A$.

  3. $A$ is true and $R$ is false.

  4. $A$ is false but $R$ is true.

Show answer
Correct Option: B
Explanation:

$a^0=1\,\,(\text{Where} \,a\ne 0)$  

$\therefore $   Assertion is true
And $\dfrac{a^m}{a^n}=a^{m-n}$
2nd statement is also true, but not the correct explanation of first statement.