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Re: querries

  • To: mathgroup at
  • Subject: [mg119576] Re: querries
  • From: Bob Hanlon <hanlonr at>
  • Date: Fri, 10 Jun 2011 06:40:36 -0400 (EDT)
  • Reply-to: hanlonr at

Because it isn't always true. For example,

((-2)*(-8))^(1/2) - (-2)^(1/2)*(-8)^(1/2)


Read help for PowerExpand

(mu*lambda)^b - (mu^b)*(lambda^b) // PowerExpand


(mu*lambda)^b - (mu^b)*(lambda^b) // Simplify[#, {mu > 0, lambda > 0}] &


Bob Hanlon

---- "Savits wrote: 

1)       Why does Mathematica not give the value 0 to the expression
(mu*lambda)^(b) - (mu^b)*(lambda^b)? It seems to treat the expression (mu*lambda) as a new variable. I used the greek letters from the palette in the actual expression.

2)       Many times I cannot get Mathematica to simplify expressions involving the exponential functions, e.g., it won't combine the arguments of the Exp functions. Is there a way to force this?

Tom Savits
Statistics Department
University of Pittsburgh

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