How do you solve #2 ln (7x) = 4#?

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Answer 1 · 2018-08-09T12:05:37

#:.x=e^2/7#

Here ,

#2ln(7x)=4# #where , x in RR^+#

#:.ln(7x)^2=4...to[becauselna^n=nlna]#

#:.(7x)^2=e^4....to[becauselny=k<=>y=e^k ]#

#:.7x=e^2#

#:.x=e^2/7#

Check :

#LHS=2ln(7x)=2ln(7*e^2/7)=2lne^2=4lne=4(1)=4#