Question #73286

1 Answer
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))