If 
prove that
We have, xm.yn = (x+y) m+n
Taking log on both sides, we get
log (xm.yn) = log (x+y) m+n
⇒ m log x + n log y = (m+n) log (x+y)
Differentiating both sides w.r.t x, we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Hence proved.
80
If prove that
We have, xm.yn = (x+y) m+n
Taking log on both sides, we get
log (xm.yn) = log (x+y) m+n
⇒ m log x + n log y = (m+n) log (x+y)
Differentiating both sides w.r.t x, we get
⇒
⇒
⇒
⇒
⇒
⇒
Hence proved.