Answer:
[tex]P(t)=\frac{5}{2} \cdot (\sqrt{2})^t[/tex]
Step-by-step explanation:
[tex]P(2)=5[/tex] meants when [tex]t=2[/tex], that the value for [tex]P(t)[/tex] is 5.
So this gives us this equation:
[tex]5=P_0 \cdot a^2[/tex]
[tex]P(4)=10[/tex] meants when [tex]t=4[/tex], that the value for [tex]P(t)[/tex] is 10.
So this gives us this equation:
[tex]10=P_0 \cdot a^4[/tex]
So I take equation 2 and divide it be equation 1 I get:
[tex]\frac{10}{5}=\frac{P_0 \cdot a_4}{P_0 \cdot a_2}[/tex]
Simplifying:
[tex]2=a^2[/tex]
Since the base for an exponential function can't be negative then [tex]a=\sqrt{2}[/tex].
So plugging into one of my equations I began with gives me an equation to solve for the initial value,[tex]P_0[/tex]:
[tex]5=P_0 \cdot (\sqrt{2})^2[/tex]
[tex]5=P_0 \cdot 2[/tex]
Divide both sides by 2:
[tex]\frac{5}{2}=P_0[/tex]
The function is:
[tex]P(t)=\frac{5}{2} \cdot (\sqrt{2})^t[/tex]