Tag: laws of exponents and powers

Questions Related to laws of exponents and powers

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.

The value of $\cfrac { { 2 }^{ 2n-2 } }{ { 2 }^{ n(n-1) } }-\cfrac { { 8 }^{ n-1 } }{ { 2 }^{ (n-1)(n+1) } } $ will be

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

  1. $2$

  2. $0$

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

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

Show answer
Correct Option: B
Explanation:
Given,

$\dfrac{2^{2n-2}}{2^{n(n-1)}} - \dfrac{8^{n-1}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^{2(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^{3(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)}2^{(n+1)}}$

$=\dfrac{2^2}{2^n}-\dfrac{2^3}{2^{n+1}}$

$=\dfrac{2^2}{2^n}-\dfrac{2^3}{2^n \times 2^1}$

$=\dfrac{2^2}{2^n}-\dfrac{2^2}{2^n}$

$=0$

Find the sum of all values of $x$, so that $16^{\left(x^{2}+3x-1\right)}=8^{\left(x^{2}+3x+2\right)}$.

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

  1. $0$

  2. $3$

  3. $-3$

  4. $-5$

Show answer
Correct Option: C
Explanation:

Given, $16^{(x^2 + 3x -1)} = 8^{(x^2 + 3x + 2)}$


As, 16 = $2^4 and \ \ \ 8 = 2^3$


$2^{4(x^2 + 3x -1)} = 2^{3(x^2 + 3x + 2)}$

So, we can write 

${4(x^2 + 3x -1)} = {3(x^2 + 3x + 2)}$

${4x^2 + 12x - 4} = {3x^2 + 9x + 6}$

${4x^2 + 12x - 4} - {(3x^2 + 9x + 6)} = 0$
$4x^2 + 12x - 4 - 3x^2 - 9x - 6 = 0$

$x^2 + 3x - 10  = 0$

$x^2 + 5x - 2x - 10  = 0$

$x(x + 5) - 2(x + 5)  = 0$

$(x - 2)(x + 5)  = 0$

So, $x = 2, - 5$

Sum of values of $x = 2 + (-5) = -3$

If  $n$  is a natural number, then  $4 ^ { n } - 3 ^ {  n }$  ends with a digit  $x.$  The number of possible values of  $x$  is

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

  1. $3$

  2. $8$

  3. $5$

  4. $6$

Show answer
Correct Option: A