IB Mathematics SL 1.7 Laws of exponents with rational exponents AA SL Paper 2- Exam Style Questions- New Syllabus

(a)
One year is 12 months, so substitute \(t=12\).

\(F(12)=1850\times1.105^{12}\)

\(F(12)=6130.827\ldots\)

Therefore, the predicted population after one year is approximately \(6130\) frogs to three significant figures.

✅ Answer: \(6130\) frogs

(b)
The population must be at least \(35\,000\), so:

\(1850\times1.105^x\geq35\,000\)

\(1.105^x\geq\dfrac{35\,000}{1850}\)

Taking natural logarithms:

\(x\ln(1.105)\geq\ln\left(\dfrac{35\,000}{1850}\right)\)

\(x\geq\dfrac{\ln\left(\frac{35\,000}{1850}\right)}{\ln(1.105)}\)

\(x\geq29.447\ldots\)

Since \(x\) represents a number of complete months, the first whole month for which the population is at least \(35\,000\) is 30.

✅ Answer: \(x=30\) months

(c)
Rewrite \(1.105^t\) using \(a^t=e^{t\ln a}\).

\(1.105^t=e^{t\ln(1.105)}\)

Therefore:

\(F(t)=1850e^{t\ln(1.105)}\)

Comparing this with \(F(t)=1850e^{kt}\):

✅ Answer: \(k=\ln(1.105)\)

(d)
The rate of population growth is given by \(F'(t)\).

Using \(F(t)=1850e^{kt}\):

\(F'(t)=1850ke^{kt}\)

Since \(k=\ln(1.105)\):

\(F'(15)=1850\ln(1.105)\times1.105^{15}\)

\(F'(15)=825.911\ldots\)

✅ Answer: Approximately \(826\) frogs per month

(e)
Substitute \(t=15\) and \(G(15)=6995\).

\(6995=\dfrac{35\,000}{1+Ae^{-0.0998(15)}}\)

\(1+Ae^{-1.497}=\dfrac{35\,000}{6995}\)

\(Ae^{-1.497}=\dfrac{35\,000}{6995}-1\)

\(A=e^{1.497}\left(\dfrac{35\,000}{6995}-1\right)\)

\(A=17.889\ldots\)

✅ Answer: \(A\approx17.9\)

(f)
For the logistic model, the growth rate is greatest at the point of inflexion, when the population is half of its carrying capacity.

The carrying capacity is \(35\,000\), so the greatest growth occurs when:

\(G(t)=17\,500\)

\(17\,500=\dfrac{35\,000}{1+Ae^{-0.0998t}}\)

\(1+Ae^{-0.0998t}=2\)

\(Ae^{-0.0998t}=1\)

Taking natural logarithms:

\(\ln A-0.0998t=0\)

\(t=\dfrac{\ln A}{0.0998}\)

Using \(A=17.889\ldots\):

\(t=\dfrac{\ln(17.889\ldots)}{0.0998}=28.899\ldots\)

✅ Answer: \(t\approx28.9\) months

(g)
The model \(G(t)\) approaches a maximum population of \(35\,000\) frogs. This represents a carrying capacity caused by limited food, space and other resources.

In contrast, \(F(t)\) increases without limit, which is not realistic over a long period.

✅ Answer: \(G(t)\) includes a carrying capacity, whereas \(F(t)\) grows without bound.