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
Simplify the following
`(4ab^2(-5ab^3))/(10a^2b^2)`
If a = 3 and b = -2, find the values of :
aa + bb
Prove that:
`(x^a/x^b)^(a^2+ab+b^2)xx(x^b/x^c)^(b^2+bc+c^2)xx(x^c/x^a)^(c^2+ca+a^2)=1`
Solve the following equation for x:
`2^(5x+3)=8^(x+3)`
If `a=xy^(p-1), b=xy^(q-1)` and `c=xy^(r-1),` prove that `a^(q-r)b^(r-p)c^(p-q)=1`
Show that:
`(x^(a^2+b^2)/x^(ab))^(a+b)(x^(b^2+c^2)/x^(bc))^(b+c)(x^(c^2+a^2)/x^(ac))^(a+c)=x^(2(a^3+b^3+c^3))`
Write the value of \[\left\{ 5( 8^{1/3} + {27}^{1/3} )^3 \right\}^{1/4} . \]
If a, m, n are positive ingegers, then \[\left\{ \sqrt[m]{\sqrt[n]{a}} \right\}^{mn}\] is equal to
If \[\frac{3^{5x} \times {81}^2 \times 6561}{3^{2x}} = 3^7\] then x =
If \[\sqrt{2} = 1 . 4142\] then \[\sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}}\] is equal to
