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Student Support Forum: 'solving inverse functions' topicStudent Support Forum > General > "solving inverse functions"

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Author Comment/Response
Bill Simpson
09/19/12 3:48pm

This does not eliminate the warnings

In[1]:= Reduce[1/2Log[x]+1/2x==y,x,Reals]

Out[1]= 0==2*y - Log[ProductLog[E^(2*y)]] - ProductLog[E^(2*y)] && x==ProductLog[E^(2*y)]

In[2]:= FullSimplify[%]

Out[2]= 2*y==Log[E^(2*y)] && x==ProductLog[E^(2*y)]

But for y==1/2 it does show x==1

In[3]:= FullSimplify[ProductLog[E^(2y)/.y->1/2]]

Out[3]= 1



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Subject (listing for 'solving inverse functions')
Author Date Posted
solving inverse functions hetanquary 09/19/12 2:27pm
Re: solving inverse functions Bill Simpson 09/19/12 3:48pm
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