1. growth and decay

I need help finding the formula for P(t)!!

A population P=P(t) grows exponentially with P(3)=60 and P(6)=90. Find the formula for P(t), and find the population when t=10.

I know how to do this but only when P(0) is given, which it is not in this case. Please give some hints = )

2. Hello, redpanda11!

A population $\displaystyle P(t)$ grows exponentially with: $\displaystyle P(3)=60$ and $\displaystyle P(6)=90$
Find the formula for $\displaystyle P(t)$
The function has the form: .$\displaystyle P(t) \;=\;ae^{bt}$

From $\displaystyle P(3) = 60$, we have: .$\displaystyle ae^{3b} \:=\:60$ . [1]

From $\displaystyle P(6)=90$, we have: .$\displaystyle ae^{6b} \:=\:90$ . [2]

Divide [2] by [1]: .$\displaystyle \frac{ae^{6b}}{ae^{3b}} \:=\:\frac{90}{60}\quad\Rightarrow\quad e^{3b} \:=\:1.5\quad\Rightarrow\quad 3b \:=\:\ln(1.5)$

.Hence: .$\displaystyle \boxed{b \:=\:\frac{1}{3}\ln(1.5)}$

The function (so far) is: .$\displaystyle P(t) \;=\;ae^{\left(\frac{1}{3}\ln15\right)t} \;=\;a\left(e^{\ln15}\right)^{\frac{1}{3}t}\quad\R ightarrow\quad P(t)\:=\:a(1.5)^{\frac{1}{3}t}$

From $\displaystyle P(3) = 60$, we have: .$\displaystyle a(1.5)^1 \:=\:60\quad\Rightarrow\quad 1.5a \:=\:60\quad\Rightarrow\quad\boxed{a \:=\:40}$

Therefore: .$\displaystyle {\color{blue}\boxed{P(t) \;=\;40(1.5)^{\frac{1}{3}t}}}$