Question

The same exponential growth function can be written in the forms y() - yo e t, y(t)- yo(1 r, and y) yo2 2. Write k as a function of r, r as a function of T2, and T2 as a function of k Write k as a function of r. k(r)-

Answer 1

`(a)\ k(r) = \ln(1+r)`

`(b)\ r(T_2) = 2^(1/T_2)-1`

`(c)\ T_2(k) = (\ln(2))/(k)`

Explanation:

Given

`y(t) = y_0e^(kt)`

`y(t) = y_0(1+r)^t`

`y(t) = y_02^(t/T_2)`

Solving (a): k(r)

Equate
`y(t) = y_0e^(kt)` and
`y(t) = y_0(1+r)^t`

`y_0e^(kt) = y_0(1+r)^t`

Cancel out common terms

`e^(kt) = (1+r)^t`

Take ln of both sides

`\ln(e^(kt)) = \ln((1+r)^t)`

Rewrite as:

`kt\ln(e) = t\ln(1+r)`

Divide both sides by t

`k = \ln(1+r)`

Hence:

`k(r) = \ln(1+r)`

Solving (b): r(T2)

Equate
`y(t) = y_02^(t/T_2)` and
`y(t) = y_0(1+r)^t`

`y_0(1+r)^t = y_02^(t/T_2)`

Cancel out common terms

`(1+r)^t = 2^(t/T_2)`

Take t th root of both sides

`(1+r)^(t*1/t) = 2^(t/T_2*1/t)`

`1+r = 2^(1/T_2)`

Make r the subject

`r = 2^(1/T_2)-1`

Hence:

`r(T_2) = 2^(1/T_2)-1`

Solving (c): T2(k)

Equate
`y(t) = y_02^(t/T_2)` and
`y(t) = y_0e^(kt)`

`y_02^(t/T_2) = y_0e^(kt)`

Cancel out common terms

`2^(t/T_2) = e^(kt)`

Take ln of both sides

`\ln(2^(t/T_2)) = \ln(e^(kt))`

Rewrite as:

`(t)/(T_2) * \ln(2) = kt\ln(e)`

`(t)/(T_2) * \ln(2) = kt*1`

`(t)/(T_2) * \ln(2) = kt`

Divide both sides by t

`(1)/(T_2) * \ln(2) = k`

Cross multiply

`kT_2 = \ln(2)`

Make T2 the subject

`T_2 = (\ln(2))/(k)`

Hence:

`T_2(k) = (\ln(2))/(k)`