Answer:
[tex](a)\ G(t) = 100 *e^{0.1823t}[/tex]
[tex](b)\ t = 6[/tex]
[tex](c)\ t = 1.7[/tex]
Step-by-step explanation:
Given
[tex]G_0 = 100[/tex] --- initial
[tex]G(1) = 120[/tex] --- after 1 year
[tex]r \to rate[/tex]
Solving (a): The expression for g
Since the rate is constant, the distribution of G follows:
[tex]G(t) = G_0 * e^{rt}[/tex]
[tex]G(1) = 120[/tex] implies that:
[tex]G(t) = G_0 * e^{rt}[/tex]
[tex]120 = G_0 * e^{r*1}[/tex]
Substitute [tex]G_0 = 100[/tex]
[tex]120 = 100 * e^{r[/tex]
Divide both sides by 100
[tex]1.2 = e^{r[/tex]
Take natural logarithm of both sides
[tex]\ln(1.2) = \ln(e^r)[/tex]
[tex]0.1823 = r[/tex]
[tex]r = 0.1823[/tex]
So, the expression for G is:
[tex]G(t) = G_0 * e^{rt}[/tex]
[tex]G(t) = 100 *e^{0.1823t}[/tex]
Solving (b): t when G(t) = 300
We have:
[tex]G(t) = 100 *e^{0.1823t}[/tex]
[tex]300 = 100 *e^{0.1823t}[/tex]
Divide both sides by 100
[tex]3 = e^{0.1823t}[/tex]
Take natural logarithm
[tex]\ln(3) = \ln(e^{0.1823t})[/tex]
[tex]1.099 = 0.1823t[/tex]
Solve for t
[tex]t = \frac{1.099}{0.1823}[/tex]
[tex]t = 6[/tex] --- approximated
Solving (c): When there will be no grass
Reduction at a rate of 80 tons per year implies that:
[tex]G(t) = 100 *e^{0.1823t}- 80t[/tex]
To solve for t, we set G(t) = 0
[tex]0 = 100 *e^{0.1823t}- 80t\\[/tex]
Rewrite as
[tex]80t = 100 *e^{0.1823t}[/tex]
Divide both sides by 100
[tex]0.8t = e^{0.1823t}[/tex]
Take natural logarithm of both sides
[tex]\ln( 0.8t) = \ln(e^{0.1823t})[/tex]
[tex]\ln( 0.8t) = 0.1823t[/tex]
Plot the graph of: [tex]\ln( 0.8t) = 0.1823t[/tex]
[tex]t = 1.7[/tex]
