Tag: power and exponent

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Questions Related to power and exponent

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

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?

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $2$
  2. $0$
  3. $\dfrac {1}{2}$
  4. $\dfrac {1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

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

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

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

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

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

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

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

$=0$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

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

  1. $0$
  2. $3$
  3. $-3$
  4. $-5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $x = 2 + (-5) = -3$