(a^m)*(a^n) = a^(m+n)
Proof
a^m = a*a*a*....a[here m times a is multiplied]
a^n = a*a*a*...a[here n times a is multiplied]
(a^m)*(a^n) = {a*a*a*....a[ m times a]}*{a*a*a*...a[ n times a]}
= a*a*a*.........*a[m times a* n times a = (m+n) times a]
= a^(m+n)
hence prooved
a^{m} * a^{n}=a^{m+n}
LHS =RHS
forward proof, LHS=RHS...............................1
backward proof, RHS=LHS..............................2
using 1,
a^{m} = a × a × ... × a →m of these
a^{n}= a × a × ... × a →n of these
a^{m}× a^{n} = a × a × ... × a × a × a × ... × a →(m+n) of these
= a^{(m+n)}
Let us consider m=5, n=4
therefore,
a^{m}=a^{5}
= a x a x a x a x a
a^{n}=a^{4}
= a x a x a x a
therefore,
a^{5} x a^{4} = a x a x a x a x a x a x a x a x a
= a^{9}
Which shows m+n = 5+4 = 9
hence,
a^{m} x a^{n} = a^{m+n}
given problem is a^m*a^n=a^(m+n)
let us take L.H.S => a^m*a^n
the base is same for two values.so base is same ,the powers will be added
now we can do that, it will become
=> a^(m+n)
now, L.H.S=R.H.S