Question: A bacterial population obeys $P = 1000 e^{0.05t}$, where $P$ is the population and $t$ is time in hours. Find $t$ when $P = 2000$.

Solution:

$$\begin{aligned} 2000 &= 1000 e^{0.05t} \\ 2 &= e^{0.05t} \\ \ln 2 &= 0.05t \\ t &= \frac{\ln 2}{0.05} \\ &\approx \boxed{13.86 \text{ hours}} \end{aligned}$$

Doubling time: like half-life, doubling time depends only on the rate $k = 0.05$, not the starting value $P_0 = 1000$. Same population would still take 13.86 hours to double from 5000 to 10000.

Question: Sketch the graph of $y = \ln(x - 2) + 1$, stating its vertical asymptote and any axis intercepts.

Solution:

Start with $y = \ln x$ (asymptote $x = 0$, passes through $(1, 0)$), then apply transformations:

• Replace $x$ with $x - 2$: shifts curve 2 units right. Asymptote moves to $x = 2$.
• Add 1 to $y$: shifts curve 1 unit up. Point $(1, 0)$ on $\ln x$ maps to $(3, 1)$ on the new curve.

Vertical asymptote: $x = 2$

$x$-intercept (where $y = 0$):

$$\begin{aligned} \ln(x - 2) + 1 &= 0 \\ \ln(x - 2) &= -1 \\ x - 2 &= e^{-1} \\ x &= 2 + e^{-1} \approx 2.37 \end{aligned}$$

No $y$-intercept because the domain is $x > 2$ (so $x = 0$ is outside the domain).

Sketch: log curve with asymptote at $x = 2$, passing through $(2 + 1/e, 0)$ and $(3, 1)$, increasing for $x > 2$.

Question: The graph of $y = \lg x$ is drawn for $0 < x \le 4$. By drawing a suitable straight line on the same axes, solve the equation $x + \lg x = 2$, and state the number of solutions.

Step 1. Rearrange until the drawn curve stands alone on one side:

$$x + \lg x = 2 \implies \lg x = 2 - x$$

Step 2. The left side is the curve already on the axes; the right side is the straight line to draw:

$$y = 2 - x \qquad (\text{gradient } -1,\; y\text{-intercept } 2)$$

Step 3. Solutions of $x + \lg x = 2$ are the $x$-coordinates of the intersections. Locate them:

• At $x = 1$: curve $\lg 1 = 0$, line $2 - 1 = 1$ — line is above.
• At $x = 2$: curve $\lg 2 \approx 0.30$, line $2 - 2 = 0$ — curve is above.

So they cross between $x = 1$ and $x = 2$.

Step 4. $y = \lg x$ is always increasing and $y = 2 - x$ is always decreasing, so they can cross only once. Reading off the graph: one solution, $x \approx 1.76$.

⚠ Watch out: The line must come from making $\lg x$ exactly the subject — a sign slip like drawing $y = x - 2$ solves a different equation, and $y = 2 - \lg x$ is not even a straight line in $x$.

Question: The variables $x$ and $y$ are related by $y = A x^n$, where $A$ and $n$ are constants. The straight-line graph of $\lg y$ against $\lg x$ has gradient $0.5$ and $y$-intercept $1.2$. Find $A$ and $n$.

Solution:

Take $\lg$ of both sides of $y = A x^n$:

$$\lg y = \lg A + n \lg x$$

Compare with $Y = mX + c$ (where $Y = \lg y$, $X = \lg x$):

• Gradient $m = n$
• $y$-intercept $c = \lg A$

From the given values:

$$n = 0.5, \qquad \lg A = 1.2 \implies A = 10^{1.2} \approx \boxed{15.85}$$ $$\boxed{A \approx 15.85, \quad n = 0.5}$$

Strategy: identify how to take logs to convert the model into linear form, then read constants off the gradient/intercept.

Question: The variables are related by $y = K \cdot a^x$. The graph of $\ln y$ against $x$ is a straight line with gradient $0.6$ and $\ln y$-intercept $2$. Find $K$ and $a$.

Solution:

Take $\ln$ of both sides:

$$\ln y = \ln K + x \ln a$$

Compare with $Y = mX + c$ (where $Y = \ln y$, $X = x$):

• Gradient $m = \ln a$
• Intercept $c = \ln K$

From the given values:

$$\begin{aligned} \ln a &= 0.6 \implies a = e^{0.6} \approx 1.82 \\ \ln K &= 2 \implies K = e^2 \approx 7.39 \end{aligned}$$ $$\boxed{K \approx 7.39, \quad a \approx 1.82}$$

Question: Without using a calculator, evaluate $\log_3 81 - \log_3 9 + \log_3 \tfrac{1}{3}$.

Key log laws:

$$\log_a (xy) = \log_a x + \log_a y, \quad \log_a \tfrac{x}{y} = \log_a x - \log_a y, \quad \log_a (x^n) = n \log_a x$$

Solution:

Each term is a power of 3:

$$\begin{aligned} \log_3 81 &= \log_3 3^4 = 4 \\ \log_3 9 &= \log_3 3^2 = 2 \\ \log_3 \tfrac{1}{3} &= \log_3 3^{-1} = -1 \end{aligned}$$

Combine:

$$4 - 2 + (-1) = \boxed{1}$$

Question: Evaluate $\log_4 32$, leaving the answer as a fraction.

Change of base formula:

$$\log_a b = \frac{\log_c b}{\log_c a} \quad \text{for any base } c$$

Solution:

Both 4 and 32 are powers of 2: $4 = 2^2$, $32 = 2^5$. Use base 2:

$$\log_4 32 = \frac{\log_2 32}{\log_2 4} = \frac{5}{2}$$ $$\boxed{\log_4 32 = \tfrac{5}{2}}$$

Strategy: when the argument and base share a common base $c$, change to base $c$ and the calculation reduces to dividing two integers.

Question: Show that $\log_a b \cdot \log_b c = \log_a c$.

Solution:

Apply change of base to both factors using base $a$:

$$\begin{aligned} \log_a b \cdot \log_b c &= \log_a b \cdot \frac{\log_a c}{\log_a b} \\ &= \log_a c \quad \text{(shown)} \end{aligned}$$

Reading: this is sometimes called the "chain rule for logs" — convert through a middle base, the middle terms cancel.

Question: Sound intensity level is measured in decibels using $L = 10 \lg \dfrac{I}{I_0}$, where $I_0 = 10^{-12}$ W/m². A jackhammer produces $L = 110$ dB. Find $I$.

Solution:

$$\begin{aligned} 110 &= 10 \lg \frac{I}{10^{-12}} \\ 11 &= \lg \frac{I}{10^{-12}} \\ \frac{I}{10^{-12}} &= 10^{11} \\ I &= 10^{11} \times 10^{-12} = 10^{-1} = \boxed{0.1 \text{ W/m}^2} \end{aligned}$$

Reading: every 10 dB increase corresponds to 10× the intensity. So 110 dB is $10^{11}$ times more intense than the threshold of hearing.

Question: Solve $7^{x+1} = 100$, giving your answer to 3 significant figures.

Solution:

Take $\lg$ (log base 10) of both sides:

$$\begin{aligned} \lg(7^{x+1}) &= \lg 100 \\ (x+1) \lg 7 &= 2 \\ x + 1 &= \frac{2}{\lg 7} \\ x &= \frac{2}{\lg 7} - 1 \\ &\approx 2.366 - 1 \\ &\approx \boxed{1.37} \end{aligned}$$

Question: Solve $3^{2x} - 4 \cdot 3^x + 3 = 0$.

Solution:

Let $u = 3^x$. Note $3^{2x} = u^2$:

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

Back-substitute:

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

This is a Logarithms-topic exercise but the substitution trick is identical to Indices sg660. Both topics overlap here.

Question: Solve $\log_2 x + \log_2 (x - 3) = 2$.

Solution:

Combine using product law:

$$\begin{aligned} \log_2 \left[x(x - 3)\right] &= 2 \\ x(x - 3) &= 2^2 \\ x^2 - 3x - 4 &= 0 \\ (x - 4)(x + 1) &= 0 \\ x &= 4 \text{ or } x = -1 \end{aligned}$$

Check domain: $\log x$ requires $x > 0$ AND $\log(x - 3)$ requires $x > 3$.

→ Reject $x = -1$.

$$\boxed{x = 4}$$

⚠️ Always check domain after solving — log equations have implicit positivity constraints that the algebra doesn't enforce.

Question: Solve $(\log_2 x)^2 - 5 \log_2 x + 6 = 0$.

Solution:

Let $u = \log_2 x$:

$$\begin{aligned} u^2 - 5u + 6 &= 0 \\ (u - 2)(u - 3) &= 0 \\ u &= 2 \text{ or } u = 3 \end{aligned}$$

Back-substitute:

$$\begin{aligned} \log_2 x = 2 &\implies x = 2^2 = 4 \\ \log_2 x = 3 &\implies x = 2^3 = 8 \end{aligned}$$ $$\boxed{x = 4 \text{ or } x = 8}$$

Question: Solve the simultaneous equations

$$\log_2 x + \log_2 y = 5, \qquad \log_2 x - \log_2 y = 1$$

Step 1. Treat the logs themselves as the unknowns: let $u = \log_2 x$ and $v = \log_2 y$. The system becomes linear:

$$u + v = 5, \qquad u - v = 1$$

Step 2. Add the two equations to eliminate $v$:

$$2u = 6 \implies u = 3$$

Subtract them to eliminate $u$:

$$2v = 4 \implies v = 2$$

Step 3. Convert each log statement back to index form, using $\log_a x = k \iff x = a^k$:

$$\log_2 x = 3 \implies x = 2^3 = 8, \qquad \log_2 y = 2 \implies y = 2^2 = 4$$ $$\therefore \; x = 8, \quad y = 4$$

Step 4. Check in the originals: both arguments are positive ($8 > 0$, $4 > 0$), and

$$\log_2 8 + \log_2 4 = 3 + 2 = 5 \;✓ \qquad \log_2 8 - \log_2 4 = 3 - 2 = 1 \;✓$$

⚠ Watch out: If you instead combine the logs first ($xy = 32$ and $\tfrac{x}{y} = 2$, so $x^2 = 64$), the algebra offers $x = \pm 8$ — you must reject $x = -8,\; y = -4$ because $\log_2 x$ and $\log_2 y$ need positive arguments in the original equations.