Question: On the same axes, sketch $y = e^x$ and $y = e^{-x}$, labelling the $y$-intercept and asymptote of each.

Solution:

$y = e^x$ (exponential growth):

• $y$-intercept: $e^0 = 1$, point $(0, 1)$
• Horizontal asymptote: $y = 0$ (as $x \to -\infty$)
• Strictly increasing, passes through $(1, e)$

$y = e^{-x}$ (exponential decay):

• $y$-intercept: $e^0 = 1$, point $(0, 1)$ (same as above)
• Horizontal asymptote: $y = 0$ (as $x \to +\infty$)
• Strictly decreasing, mirror image of $y = e^x$ about the $y$-axis

Both curves intersect at $(0, 1)$. Both lie strictly above the $x$-axis.

Key fact: $e^x > 0$ for all real $x$ — the curve never crosses or touches the $x$-axis.

Question: Solve $\log_{\sqrt{3}}(x - 1) = \log_3(x + 5)$.

Solution:

Convert $\log_{\sqrt{3}}$ to base 3 using change of base:

$$\log_{\sqrt{3}}(x - 1) = \frac{\log_3(x-1)}{\log_3 \sqrt{3}} = \frac{\log_3(x-1)}{1/2} = 2\log_3(x - 1)$$

The equation becomes:

$$\begin{aligned} 2\log_3(x - 1) &= \log_3(x + 5) \\ \log_3(x - 1)^2 &= \log_3(x + 5) \\ (x - 1)^2 &= x + 5 \\ x^2 - 2x + 1 &= x + 5 \\ x^2 - 3x - 4 &= 0 \\ (x - 4)(x + 1) &= 0 \end{aligned}$$

So $x = 4$ or $x = -1$.

Check domain: $\log(x - 1)$ requires $x > 1$ → reject $x = -1$.

$$\boxed{x = 4}$$

Question: Solve $9^{x+1} = 27^{2x - 1}$.

Solution:

Both bases are powers of 3: $9 = 3^2$ and $27 = 3^3$.

$$\begin{aligned} (3^2)^{x+1} &= (3^3)^{2x-1} \\ 3^{2(x+1)} &= 3^{3(2x-1)} \\ 2(x+1) &= 3(2x-1) \quad \text{(equate exponents)} \\ 2x + 2 &= 6x - 3 \\ 5 &= 4x \\ x &= \boxed{\tfrac{5}{4}} \end{aligned}$$

Strategy: if both sides can be expressed in the same base, equate exponents. No logs needed.

Question: Solve $5^{2x} = 7$, leaving your answer in the form $\dfrac{\lg a}{\lg b}$.

Solution:

Bases 5 and 7 are coprime — can't be matched. Take $\lg$ of both sides:

$$\begin{aligned} \lg(5^{2x}) &= \lg 7 \\ 2x \lg 5 &= \lg 7 \\ x &= \boxed{\frac{\lg 7}{2 \lg 5}} \end{aligned}$$

Equivalent form: $\dfrac{\lg 7}{\lg 25}$ (using $2 \lg 5 = \lg 5^2 = \lg 25$).

Question: Given that $4u^3 - 12u^2 - 15u + 4 = (4u - 1)(u^2 - 4u - 4) \cdot \text{(adjustment)}$… actually given that $4u^3 - 12u^2 - 15u + 4 = 0$ has $u = 4$ as a root, solve $4(9^x) = 12(3^x) + 15 - \dfrac{4}{3^x}$.

Solution:

Multiply the equation by $3^x$ to clear fractions:

$$4 \cdot 9^x \cdot 3^x = 12 \cdot 3^x \cdot 3^x + 15 \cdot 3^x - 4$$

Note $9^x \cdot 3^x = 3^{3x}$ and $3^x \cdot 3^x = 3^{2x}$. Let $u = 3^x$:

$$4u^3 = 12u^2 + 15u - 4 \implies 4u^3 - 12u^2 - 15u + 4 = 0$$

This is the cubic with $u = 4$ as a root. Factor:

$$4u^3 - 12u^2 - 15u + 4 = (u - 4)(4u^2 + 4u - 1) = 0$$

So $u = 4$ or $4u^2 + 4u - 1 = 0$, giving $u = \dfrac{-4 \pm \sqrt{32}}{8} = \dfrac{-1 \pm \sqrt{2}}{2}$.

Reject $u = \tfrac{-1 - \sqrt{2}}{2} < 0$ and $u = \tfrac{-1 + \sqrt{2}}{2} \approx 0.21$ → check positive.

Valid roots: $u = 4$ and $u = \tfrac{-1 + \sqrt{2}}{2}$.

Convert via logs:

$$3^x = 4 \implies x = \log_3 4 = \frac{\lg 4}{\lg 3}$$ $$3^x = \tfrac{-1 + \sqrt{2}}{2} \implies x = \log_3\!\left(\tfrac{-1 + \sqrt{2}}{2}\right) \approx -1.43$$

Question: Simplify $\dfrac{(2x^3)^4 \cdot x^{-2}}{4x^5}$.

Key index laws:

$$a^m \cdot a^n = a^{m+n}, \quad \frac{a^m}{a^n} = a^{m-n}, \quad (a^m)^n = a^{mn}$$

Solution:

$$\begin{aligned} \frac{(2x^3)^4 \cdot x^{-2}}{4x^5} &= \frac{2^4 \cdot x^{12} \cdot x^{-2}}{4x^5} \\ &= \frac{16 \cdot x^{12 + (-2)}}{4 x^5} \\ &= \frac{16 x^{10}}{4 x^5} \\ &= 4 x^{10 - 5} \\ &= \boxed{4x^5} \end{aligned}$$

Question: Solve the simultaneous equations $2^x \cdot 32^y = 1$ and $27^x = 3^{y - 12}$.

Solution:

Convert all bases to primes (2 and 3):

$$\begin{aligned} 2^x \cdot 32^y &= 1 \implies 2^x \cdot 2^{5y} = 2^0 \implies x + 5y = 0 \quad \text{(1)} \\ 27^x &= 3^{y - 12} \implies 3^{3x} = 3^{y - 12} \implies 3x = y - 12 \quad \text{(2)} \end{aligned}$$

From (1): $y = -\tfrac{x}{5}$. Substitute into (2):

$$\begin{aligned} 3x &= -\tfrac{x}{5} - 12 \\ 15x &= -x - 60 \\ 16x &= -60 \\ x &= -\tfrac{15}{4} \end{aligned}$$

Then $y = -\tfrac{1}{5}\left(-\tfrac{15}{4}\right) = \tfrac{3}{4}$.

$$\boxed{x = -\tfrac{15}{4}, \quad y = \tfrac{3}{4}}$$

Question: Solve $9^x - 4 \cdot 3^x + 3 = 0$.

Solution:

Recognise $9^x = (3^x)^2$. Let $u = 3^x$:

$$\begin{aligned} u^2 - 4u + 3 &= 0 \\ (u - 1)(u - 3) &= 0 \\ u &= 1 \text{ or } u = 3 \end{aligned}$$

Now back-substitute $u = 3^x$:

$$\begin{aligned} 3^x = 1 &\implies x = 0 \\ 3^x = 3 &\implies x = 1 \end{aligned}$$ $$\boxed{x = 0 \text{ or } x = 1}$$

Pattern recognition: whenever you see $a^{2x} \pm \cdot a^x \pm \cdot$, substitute $u = a^x$ to get a quadratic in $u$.

Question: Solve $4^x + 2^{x+1} - 8 = 0$.

Solution:

Note $4^x = (2^x)^2$ and $2^{x+1} = 2 \cdot 2^x$. Let $u = 2^x$:

$$\begin{aligned} u^2 + 2u - 8 &= 0 \\ (u + 4)(u - 2) &= 0 \\ u &= -4 \text{ or } u = 2 \end{aligned}$$

Reject $u = -4$ because $u = 2^x > 0$ for all real $x$.

Take the valid root $u = 2$:

$$2^x = 2 \implies x = \boxed{1}$$

⚠️ Pitfall: never accept a negative or zero value of $u = a^x$ — exponentials are strictly positive.

Question: A radioactive sample decays according to $m = 200 \cdot e^{-0.05t}$, where $m$ is the mass in grams and $t$ is time in days. Find the time taken for the mass to halve.

Solution:

Initial mass at $t = 0$ is $200$ g. Halving means $m = 100$:

$$\begin{aligned} 100 &= 200 \cdot e^{-0.05t} \\ \tfrac{1}{2} &= e^{-0.05t} \\ \ln \tfrac{1}{2} &= -0.05t \\ -\ln 2 &= -0.05t \\ t &= \frac{\ln 2}{0.05} \\ &\approx \boxed{13.86 \text{ days}} \end{aligned}$$

Useful identity: $\ln \tfrac{1}{a} = -\ln a$. The half-life is independent of the starting mass — a characteristic of exponential decay.

Question: The temperature of a cup of coffee is modelled by $T = 25 + 65 e^{-0.2t}$, where $T$ is °C and $t$ is in minutes.

(a) State the room temperature (limiting value as $t \to \infty$). (b) Find the time when $T = 50$°C.

Solution:

(a) As $t \to \infty$, $e^{-0.2t} \to 0$, so $T \to 25$. Room temperature is 25°C.

(b) Set $T = 50$:

$$\begin{aligned} 50 &= 25 + 65 e^{-0.2t} \\ 25 &= 65 e^{-0.2t} \\ \tfrac{25}{65} &= e^{-0.2t} \\ \ln \tfrac{5}{13} &= -0.2t \\ t &= -\frac{\ln(5/13)}{0.2} = \frac{\ln(13/5)}{0.2} \\ &\approx \boxed{4.78 \text{ min}} \end{aligned}$$