Question

Put the function P = 112(0.5)t in the form P = P0ekt. (Round your answers to four decimal places.)
P0 =
k ≈

Answer 1

It says to turn the exponential with a base 0.5 into one with a base 'e'.

\[\large P=112*(0.5)^t\]
\[\large P=P _{0}e ^{kt}\]

Answer 2

i would just use a couple data points to compare the two forms...
From the given equation with base 0.5
for t=0, P=112*1 = 112
for t=1, P=112*0.5 = 56

Use those values in the base e exponential to get values for Po and k.

Answer 3

Two data points used


\[P=P _{0}e^{kt}\]
(t,P) = (0, 112)
\[\large 112=P _{0}*e^{k*0}~~~~~so~~~~~P _{0}=112\]

(t,P)=(1,56)
\[P=112e ^{kt}\]
\[\large 56=112*e ^{k*1}\]

Answer 4

solving that k=ln(1/2) = -0.693

That is it,
\[\large P=112*(0.5)^t\]
same as
\[\large P=112*e^{-0.693*t}\]

Answer 5

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