go back to your previous thread.
http://www.mathhelpforum.com/math-he...-cake-4-u.html
I have an equation of
N(t)= 3000e^(kt) - 4500 (t-3)
Set N(t)=0, how can i find t value???
and k is given: k=(1/3)* ln(3)
Can u explain me with simple rules...and answer
Thank you
go back to your previous thread.
http://www.mathhelpforum.com/math-he...-cake-4-u.html
there is a property for log or Ln said
log y^2 = 2 log y & log e^y = y & e^log(y) = y
so
k = 1/3 ln(3) = ln( 3^(1/3))
N(t)= 3000e^(kt) - 4500 (t-3)
3000e^(kt) = 4500 (t-3)
3000e^(t ln( 3^(1/3)) = 4500 (t-3)
3000e^( ln(3^(t/3)) = 4500 (t-3)
30(3^(t/3)) = 45 (t-3)
6 3^(t/3) = 3^2 (t-3)
I think there is a mistake in this question
t/3 = 2 ...... t=6
6 = t-3 ......t=9
I think 1500 should be instead of 3000
or 13500 instead of 4500
I think