Thread: Need help!! logarithms and exp.func thx!

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

Originally Posted by mookie

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