Answer:
[tex](a)\ k(r) = \ln(1+r)[/tex]
[tex](b)\ r(T_2) = 2^{1/T_2}-1[/tex]
[tex](c)\ T_2(k) = \frac{\ln(2)}{k}[/tex]
Step-by-step explanation:
Given
[tex]y(t) = y_0e^{kt}[/tex]
[tex]y(t) = y_0(1+r)^t[/tex]
[tex]y(t) = y_02^{t/T_2}[/tex]
Solving (a): k(r)
Equate [tex]y(t) = y_0e^{kt}[/tex] and [tex]y(t) = y_0(1+r)^t[/tex]
[tex]y_0e^{kt} = y_0(1+r)^t[/tex]
Cancel out common terms
[tex]e^{kt} = (1+r)^t[/tex]
Take ln of both sides
[tex]\ln(e^{kt}) = \ln((1+r)^t)[/tex]
Rewrite as:
[tex]kt\ln(e) = t\ln(1+r)[/tex]
Divide both sides by t
[tex]k = \ln(1+r)[/tex]
Hence:
[tex]k(r) = \ln(1+r)[/tex]
Solving (b): r(T2)
Equate [tex]y(t) = y_02^{t/T_2}[/tex] and [tex]y(t) = y_0(1+r)^t[/tex]
[tex]y_0(1+r)^t = y_02^{t/T_2}[/tex]
Cancel out common terms
[tex](1+r)^t = 2^{t/T_2}[/tex]
Take t th root of both sides
[tex](1+r)^{t*1/t} = 2^{t/T_2*1/t}[/tex]
[tex]1+r = 2^{1/T_2}[/tex]
Make r the subject
[tex]r = 2^{1/T_2}-1[/tex]
Hence:
[tex]r(T_2) = 2^{1/T_2}-1[/tex]
Solving (c): T2(k)
Equate [tex]y(t) = y_02^{t/T_2}[/tex] and [tex]y(t) = y_0e^{kt}[/tex]
[tex]y_02^{t/T_2} = y_0e^{kt}[/tex]
Cancel out common terms
[tex]2^{t/T_2} = e^{kt}[/tex]
Take ln of both sides
[tex]\ln(2^{t/T_2}) = \ln(e^{kt})[/tex]
Rewrite as:
[tex]\frac{t}{T_2} * \ln(2) = kt\ln(e)[/tex]
[tex]\frac{t}{T_2} * \ln(2) = kt*1[/tex]
[tex]\frac{t}{T_2} * \ln(2) = kt[/tex]
Divide both sides by t
[tex]\frac{1}{T_2} * \ln(2) = k[/tex]
Cross multiply
[tex]kT_2 = \ln(2)[/tex]
Make T2 the subject
[tex]T_2 = \frac{\ln(2)}{k}[/tex]
Hence:
[tex]T_2(k) = \frac{\ln(2)}{k}[/tex]
