# How do you find the exponential function, n(t)=n_oe^(kt) that satisfies the conditions n(0)=3 and n(7)=21?

Feb 12, 2015

$n \left(t\right) = {n}_{0} {e}^{k t}$
if ${n}_{0} = 3$ and $k = \ln \frac{7}{7}$
then
n(0) = 3 e^(ln(7)/7)(0)) = 3 e^0 = 3 (1) = 3 as required
and
n(7) = 3 e^(ln(7)/7)(7)) = 3 e^(ln(7)) = 3 * 7 = 21 as required

How we got there :
We are given that:
$n \left(0\right) = 3$ and
$n \left(0\right) = {n}_{0} {e}^{\left(k\right) \left(0\right)}$ after substituting 0 for $t$
or
$n \left(0\right) = {n}_{0} {e}^{0} = {n}_{0} \cdot \left(1\right) = {n}_{0}$
So,
${n}_{0} = n \left(0\right) = 3$

Now we have
$n \left(7\right) = 21$ (given)
and
$n \left(7\right) = 3 \cdot {e}^{\left(k\right) \left(7\right)}$
so
${e}^{\left(k\right) \left(7\right)} = 7$

Since $\ln \left({e}^{p}\right) = p$
if we take the natural log of both sides
$\ln \left({e}^{\left(k\right) \left(7\right)}\right) = \ln \left(7\right)$
becomes
$\left(k\right) \left(7\right) = \ln \left(7\right)$
or
$k = \ln \frac{7}{7}$

and the original equation:
$n \left(t\right) = {n}_{0} {e}^{k t}$
becomes
$n \left(t\right) = 3 {e}^{\left(\frac{\ln \left(7\right)}{7}\right) \left(t\right)}$