If
prove that
(CBSE 2014,2017)
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
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Hence proved.
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(CBSE 2014)

(CBSE 2014)