Advertisements
Advertisements
Question
State the quotient law of exponents.
Advertisements
Solution
The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. If a is a non-zero real number and m, n are positive integers, then `a^m/a^n = a^(m-n)`
We shall divide the proof into three parts
(i) when m>n
(ii) when m = n
(iii) when m < n
Case 1
When m > n
We have
\[\frac{a^m}{a^n} = \frac{a \times a \times a . . . .\text { to m factors }}{a \times a \times a . . . . \text { to n factors }}\]
\[\frac{a^m}{a^n} = a \times a \times a . . . . to (m - n) \text { factors }\]
\[\frac{a^m}{a^n} = a^{m - n}\]
Case 2
When m = n
We get
`a^m/a^n = a^m/a^m`
Cancelling common factors in numerator and denominator we get,
`a^m/a^n = 1`
By definition we can write 1 as a°
`a^m/a^n = a^(m-m)`
`a^m/a^n = a^(m-n)`
Case 3
When m < n
In this case, we have
`a^m/a^n = 1/(axx axx a ....(n-m))`
`a^m/a^n = 1/(a^(n-m))`
`a^m/a^n = a^-(n-m)`
`a^m/a^n = a^(m-n)`
Hence `a^m/a^n = a^(m-n)`, whether m < n, m = n or,m > n
APPEARS IN
RELATED QUESTIONS
Solve the following equation for x:
`4^(2x)=1/32`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrt2/sqrt3)^5(6/7)^2`
Show that:
`(x^(a-b))^(a+b)(x^(b-c))^(b+c)(x^(c-a))^(c+a)=1`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Show that:
`(a^(x+1)/a^(y+1))^(x+y)(a^(y+2)/a^(z+2))^(y+z)(a^(z+3)/a^(x+3))^(z+x)=1`
Show that:
`((a+1/b)^mxx(a-1/b)^n)/((b+1/a)^mxx(b-1/a)^n)=(a/b)^(m+n)`
The value of \[\left\{ 2 - 3 (2 - 3 )^3 \right\}^3\] is
If (23)2 = 4x, then 3x =
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
The value of 64-1/3 (641/3-642/3), is
