How do you solve #6.2 e^(0.64t)=3e^(t+1)#?

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Answer 1 · 2017-05-05T10:15:42

See below.

#6.2 e^(0.64t)=3e^(t+1)#

so

#6.2/3=e^((1-0.64)t+1)#

or

#(1-0.64)t+1=log_e(6.2/3)#

then

#t=(log_e(6.2/3)-1)/(1-0.64) approx -0.761286#