P= 12e^(-0.02t)
What is the growth rate?
I thought it would be -1.98% because I punched in the e^-0.02 part and got .98 something and then that minus 1 is r so I got -1.98.
for exponential growth (continuous compounding)
amount = original * e^rt
rate = -.02 or -2 percent
for example if you have ten dollars
then after one year you will have
10 e^-.02
= 10(.980298) = nine dollars and eighty cents
so each year multiply by .98
NOT 1.98
BUT that is an approximation
It is really .980298 etc
but growth rate is negative 2%
Thank you!
To find the growth rate in this equation, you need to use the exponential equation form P = Pe^rt, where P is the initial value, r is the growth rate, and t is the time.
In this case, you have the equation P = 12e^(-0.02t). Comparing it to the exponential equation form, you can see that the initial value (P) is 12.
To determine the growth rate (r), you need to isolate it. Let's rearrange the equation:
P = 12e^(-0.02t)
Divide both sides by 12 to isolate e^(-0.02t):
P/12 = e^(-0.02t)
Now, take the natural logarithm (ln) of both sides to undo the e^(-0.02t):
ln(P/12) = ln(e^(-0.02t))
Using the properties of logarithms (ln(a^b) = b * ln(a)), the right side simplifies:
ln(P/12) = -0.02t
The equation is now in the form ln(P/12) = rt. Comparing it with the general form ln(a) = r, you can see that r equals -0.02 (since -0.02t is equivalent to rt).
Therefore, the growth rate is -0.02, or in percentage form, -2%. In your calculation, you mistakenly took -0.02 and converted it to -1.98%, which is close but not accurate.