Question #73286

1 Answer
Tamir E.
Nov 16, 2017

For #(x>4) in F:#
#e^(-ln(x-4)-x)=1/((x-4)(e^(x))#

Explanation:

laws:
1) #a^(log_ab)=b#
2) #e^(lnb)=b#
3) #lna iff a>0#

#e^(-ln(x-4)-x)=e^(-ln(x-4))e^(-x)=1/(e^(ln(x-4)))*1/(e^(x))#

#(e^(ln(x-4))=x-4)#

#e^(-ln(x-4)-x)=...=1/((x-4)(e^(x))#

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