Solutions to Application Questions — Exponentials & Logarithms

Substitute $t=0$ into the formula. Anything to the power of $0$ is $1$, so $2^{0}=1$.

$T(0) = 20 + 160 \times 2^{0} = 20 + 160 \times 1 = 20 + 160 = \boxed{180^{\circ}\text{C}}$.

Initial temperature = 180°C

Substitute $t=10$ and $T=100$ and solve:

1. $100 = 20 + 160 \times 2^{-10k}$
2. $80 = 160 \times 2^{-10k}$  (subtract 20)
3. $\dfrac{1}{2} = 2^{-10k}$  (divide by 160)
4. $2^{-1} = 2^{-10k}$  (write LHS as a power of 2)
5. Equate indices: $-1 = -10k \Rightarrow k = \dfrac{1}{10} = 0.1$. $\checkmark$

⚠️ "Show that" — every line of working has to be present. Do not skip the equate-indices step.

Key features:

• Starting point at $(0, 180)$ (from part a).
• Horizontal asymptote at $y = 20$ (the vertical translation).
• Decreasing exponential — passes through $(10, 100)$, $(20, 60)$, $(30, 40)$, $(40, 30)$, $(50, 25)$.

Endpoint: $(0, 180)$.   Asymptote: $T = 20$.

The exponential $2^{-0.1t}$ is always positive and decreases from $1$ (at $t=0$) toward $0$ (as $t\to\infty$). So $160\cdot 2^{-0.1t}$ decreases from $160$ down to $0$ (never reaching it), and $T$ decreases from $180$ down to just above $20$.

Range: $20 < T \le 180$   (or $(20, 180]$)

$T(20) = 20 + 160 \times 2^{-2} = 20 + 160 \times \dfrac{1}{4} = 20 + 40 = \boxed{60^{\circ}\text{C}}$.

Set $T = 25$:

1. $25 = 20 + 160 \times 2^{-0.1t}$
2. $5 = 160 \times 2^{-0.1t} \Rightarrow 2^{-0.1t} = \dfrac{5}{160} = \dfrac{1}{32}$
3. $\dfrac{1}{32} = 2^{-5}$, so $2^{-0.1t} = 2^{-5} \Rightarrow -0.1t = -5$
4. $t = 50$.

50 minutes after removal

Find $t$ at each end. Same technique as part (f):

• $T = 60$: $40 = 160\cdot 2^{-0.1t} \Rightarrow 2^{-0.1t} = \tfrac{1}{4} = 2^{-2} \Rightarrow t = 20$.
• $T = 40$: $20 = 160\cdot 2^{-0.1t} \Rightarrow 2^{-0.1t} = \tfrac{1}{8} = 2^{-3} \Rightarrow t = 30$.

Window = $30 - 20 = \boxed{10}$ minutes.   10 minutes

Average rate between two times is $\dfrac{T(40)-T(20)}{40-20}$.

• $T(20) = 60$ (from part e).
• $T(40) = 20 + 160\cdot 2^{-4} = 20 + 160\cdot\tfrac{1}{16} = 20 + 10 = 30$.

Average rate $= \dfrac{30 - 60}{20} = \dfrac{-30}{20} = \boxed{-1.5\;^{\circ}\text{C/min}}$.

The negative sign shows the cake is cooling. Always include the negative sign — losing this is a common cause of mark loss.

As $t\to\infty$, $2^{-0.1t}\to 0$, so $T \to 20 + 0 = 20$.

The temperature approaches 20°C (room temperature)

The graph runs from $t=0$ (2004) to $t=10$ (2014).

Domain: $t \in [0,\,10]$

$n(0) = 120 \log_{10}(0+3) = 120\log_{10}(3) \approx 120 \times 0.4771 \approx 57.26$.

Rounded to the nearest wombat: 57 wombats

$\log_{10}$ is an increasing function, so $n$ increases on $[0,10]$. Maximum is at the right endpoint, $t=10$:

$n(10) = 120\log_{10}(13) \approx 120 \times 1.1139 \approx 133.67$.

Max ≈ 134 wombats, in 2014 ($t=10$)

$\text{avg rate} = \dfrac{n(10) - n(0)}{10 - 0} = \dfrac{133.67 - 57.26}{10} = \dfrac{76.41}{10} \approx 7.64$.

≈ 7.6 wombats per year

$n$ runs from $n(0)\approx 57.26$ up to $n(10)\approx 133.67$.

Range: $n \in [57.26,\;133.67]$   (≈ $[57,\,134]$ wombats)

1. $120 \log_{10}(t+3) = 100$
2. $\log_{10}(t+3) = \dfrac{100}{120} = \dfrac{5}{6}$
3. $t+3 = 10^{5/6} \approx 6.813$
4. $t \approx 3.813$ years.

That's 3 years and $0.813 \times 12 \approx 9.76$ months. From the start of 2004, that lands in late October 2007.

≈ October 2007 (about 3 yr 10 mo after start of 2004)

Step 1 — use $t=0$ to find $a$:

$P(0) = a \cdot b^{0} = a = 20000$, so $a = 20\,000$.

Step 2 — sub into $P(10) = 48000$:

1. $20000 \cdot b^{10} = 48000$
2. $b^{10} = \dfrac{48000}{20000} = 2.4$
3. $b = 2.4^{1/10} \approx 1.09149...$

$a = 20\,000$ and $b \approx 1.0915$ (4 dp)

End of 2020 corresponds to $t = 20$ years after 2000.

$P(20) = 20000 \cdot (1.0915)^{20}$.

Shortcut: since $b^{10} = 2.4$, we have $b^{20} = (b^{10})^{2} = 2.4^{2} = 5.76$ exactly.

$P(20) = 20000 \times 5.76 = \boxed{115\,200}$ people.

≈ 115,200 people at the end of 2020

Solve $22000 \cdot 1.10^{t} = 100000$ for $t$:

1. $1.10^{t} = \dfrac{100000}{22000} = \dfrac{100}{22} = \dfrac{50}{11}$
2. Take $\log_{10}$ of both sides: $t\log_{10}(1.10) = \log_{10}\!\left(\dfrac{50}{11}\right)$
3. $t = \dfrac{\log_{10}(50/11)}{\log_{10}(1.10)} = \dfrac{\log_{10}(50) - \log_{10}(11)}{\log_{10}(1.10)}$
4. $t \approx \dfrac{1.6990 - 1.0414}{0.04139} = \dfrac{0.6576}{0.04139} \approx 15.89$ years

So the population reaches 100,000 at $t \approx 15.89$ years, which is during 2015 (about November). Since $15.89 < 16$, the model exceeds 100,000 before 2016. $\checkmark$

$t \approx 15.89 < 16$, so passes 100,000 in 2015 (before 2016)

$N(0) = 20 \cdot 1.5^{0} = 20$.   20 goats initially

$N(5) = 20 \times 1.5^{5} = 20 \times 7.59375 \approx 151.875$.

Rounded: ≈ 152 goats

1. $20 \cdot 1.5^{t} = 100$
2. $1.5^{t} = 5$
3. $t = \log_{1.5}(5) = \dfrac{\ln 5}{\ln 1.5} = \dfrac{1.6094}{0.4055} \approx 3.969$ years

From start of 2008, $3.969$ years later $= 3$ years and $0.969 \times 12 \approx 11.6$ months. That's late December 2011.

During 2011 (specifically December 2011)

At the start of 2015, $t = 2$:

1. $152 + A \log_{10}(5(2)+1) = 173$
2. $A \log_{10}(11) = 21$
3. $A = \dfrac{21}{\log_{10}(11)} = \dfrac{21}{1.0414} \approx 20.166...$

To the nearest whole number, $A \approx 20$. $\checkmark$

With $A=20$ the model becomes $P(t) = 152 + 20\log_{10}(5t+1)$. After $t=2$ the herd will keep growing. The rate of growth at $t=2$ is:

1. $\dfrac{dP}{dt} = 20 \cdot \dfrac{5}{(5t+1)\ln 10} = \dfrac{100}{(5t+1)\ln 10}$
2. At $t=2$: $\dfrac{dP}{dt} = \dfrac{100}{11 \times 2.3026} = \dfrac{100}{25.33} \approx 3.95$.

So the herd grows by roughly 4 goats per year at this point. To hold it steady at 173, Dora needs to cull this many extra goats each year.

≈ 4 extra goats per year

Check with the discrete rate: $P(3) - P(2) = 20\log_{10}(16) - 20\log_{10}(11) = 20\log_{10}\!\left(\dfrac{16}{11}\right) \approx 3.25$. Same ball-park.

$F(0) = 20000 \times 2^{0} = 20\,000$.    $G(0) = 10000 \times 2^{0} + 1500 = 11\,500$.

$F(0) = 20\,000$ and $G(0) = 11\,500$

50% of the initial 20,000 is 10,000. Solve $F(t) = 10000$:

1. $10000 = 20000 \times 2^{-0.6t}$
2. $2^{-0.6t} = \dfrac{1}{2} = 2^{-1}$
3. $-0.6t = -1 \Rightarrow t = \dfrac{1}{0.6} = \dfrac{5}{3} \approx 1.67$ days.

$t = \tfrac{5}{3} \approx 1.67$ days

Compute $G(5)$:

$G(5) = 10000 \times 2^{-0.2(5)} + 1500 = 10000 \times 2^{-1} + 1500 = 5000 + 1500 = 6500$.

Percentage of the initial 11,500 still alive: $\dfrac{6500}{11500} \times 100\% \approx 56.5217...\%$.

$\approx 56.52\%$

Equation to solve: $20000 \times 2^{-0.6t} = 10000 \times 2^{-0.2t} + 1500$.

This cannot be rearranged to "same base", so use CAS. Test values to locate it first, then solve:

• At $t=1$: $F \approx 13195$, $G \approx 10206$, so $F > G$.
• At $t=2$: $F \approx 8706$, $G \approx 9079$, so $F < G$.
• Crossover sits between $t=1$ and $t=2$. CAS solve gives $t \approx 1.858$.

$t \approx 1.86$ days

⚠️ Always state the equation that you would put on CAS before pressing solve. The mark for "writing an equation" is often separate from the answer mark.

As $x \to \infty$, $e^{-x} \to 0$, so $f(x) \to a + b\cdot 0 = a$. The asymptote is $y = a$.

We're told the asymptote is $y = 1$, so $a = 1$.

The graph passes through $(0, 0)$, so $f(0) = 0$:

1. $f(0) = a + b \cdot e^{0} = a + b = 1 + b$
2. $1 + b = 0 \Rightarrow b = -1$.

1. $200 \cdot 2^{-t/50} + 280 = 460$
2. $200 \cdot 2^{-t/50} = 180$
3. $2^{-t/50} = 0.9$
4. $-\dfrac{t}{50} = \log_{2}(0.9) \approx -0.152$
5. $t \approx 50 \times 0.152 \approx 7.6$ months.

Answer B: 7.6 months

Key values to mark:

• $Q(0) = 200 \cdot 1 + 280 = 480$ (starting value at $t=0$).
• $Q(24) = 200 \cdot 2^{-24/50} + 280 \approx 200 \cdot 0.7165 + 280 \approx 423.3$.

The curve decreases slowly from $(0, 480)$ down to $(24, 423.3)$. Plot both functions on the same axes:

The possums start below the lizards ($P(0)=400 < Q(0)=480$) and end above them ($P(24)\approx 468.5 > Q(24)\approx 423.3$), so the curves cross exactly once.

Set $P(t) = Q(t)$ and use CAS:

$200\log_{10}(5t+100) = 200\cdot 2^{-t/50} + 280 \Rightarrow t \approx 13.71$ months.

The possums are above the lizards from $t \approx 13.71$ to $t = 24$, a duration of

$24 - 13.71 \approx 10.29 \approx \boxed{10.3}$ months.

Answer A: 10.3 months