Applications of exponentials and logarithms

a) $V(0)=900\cdot 0.85^0=900$ euros.

b) The factor is $a=0.85=1-0.15$: it loses $15\%$ each year.

c) $V(3)=900\cdot 0.85^3=900\cdot \frac{17^3}{20^3}=\frac{44217}{80}=552.7125$ euros, about $552.71$ euros.

d) Solve $900\cdot 0.85^t<300$, that is, $0.85^t<\frac{1}{3}$. With the equality,

$$t=\frac{\ln \frac{1}{3}}{\ln 0.85}\approx 6.76$$

Since $0.85<1$ the function decreases, and the value is less than $300$ for $t>6.76$. The smallest whole number is $t=7$. Check: $V(6)\approx 339.4$ and $V(7)\approx 288.5$.

a) After $8$ days half the mass remains: $M(8)=40$. Then $80{\mathrm{e}}^{-8k}=40$ and ${\mathrm{e}}^{-8k}=\frac{1}{2}$. Taking logarithms, $-8k=\ln \frac{1}{2}=-\ln 2$, so

$$k=\frac{\ln 2}{8}\approx 0.0866$$

b) $M(20)=80{\mathrm{e}}^{-20k}=80{\mathrm{e}}^{-\frac{20\ln 2}{8}}=80\cdot 2^{-5/2}=\frac{80}{4\sqrt{2}}=10\sqrt{2}\approx 14.1$ grams.

c) Solve $80{\mathrm{e}}^{-kt}=5$: ${\mathrm{e}}^{-kt}=\frac{1}{16}=2^{-4}$, so $kt=4\ln 2$ and

$$t=\frac{4\ln 2}{k}=\frac{4\ln 2}{\ln 2/8}=32\text{ days}$$

Check by thinking in half-lives: from $80$ to $5$ grams there are four halvings ($80\to 40\to 20\to 10\to 5$), that is, $4\cdot 8=32$ days.

a) At $t=0$: $20+A=90$, so $A=70$.

b) At $t=10$: $20+70{\mathrm{e}}^{-10k}=50$, hence ${\mathrm{e}}^{-10k}=\frac{30}{70}=\frac{3}{7}$. Taking logarithms, $-10k=\ln \frac{3}{7}$ and

$$k=\frac{1}{10}\ln \frac{7}{3}\approx 0.0847$$

c) Solve $20+70{\mathrm{e}}^{-kt}=30$: ${\mathrm{e}}^{-kt}=\frac{1}{7}$, so $kt=\ln 7$ and

$$t=\frac{\ln 7}{k}=\frac{10\ln 7}{\ln \frac{7}{3}}\approx 22.97\text{ minutes}$$

d) Since ${\mathrm{e}}^{-30k}={\left({\mathrm{e}}^{-10k}\right)}^3={\left(\frac{3}{7}\right)}^3=\frac{27}{343}$,

$$T(30)=20+70\cdot \frac{27}{343}=20+\frac{270}{49}=\frac{1250}{49}\approx 25.5{}^{\circ }\mathrm{C}$$

The number $20$ is the limiting value: the horizontal asymptote $y=20$, the room temperature, which the coffee tends to as $t\to +\infty$.

In the exam: make the step "isolate the exponential" clear before taking logarithms, because that step earns a mark. With the value of a power (${\mathrm{e}}^{-30k}$), use the one already calculated to avoid rounding errors, and always write the units.