trying to solve for k reach as far as here 18 = 20 + (11 - 20)e^{3k}
M math321 New member Joined Nov 18, 2010 Messages 9 Nov 19, 2010 #1 trying to solve for k reach as far as here 18 = 20 + (11 - 20)e^{3k}
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 19, 2010 #2 \(\displaystyle 18=20+(11-20)e^{3k}\) \(\displaystyle -2=-9e^{3k}\) \(\displaystyle \frac{2}{9}=e^{3k}\) Now, finish?.
\(\displaystyle 18=20+(11-20)e^{3k}\) \(\displaystyle -2=-9e^{3k}\) \(\displaystyle \frac{2}{9}=e^{3k}\) Now, finish?.
