Logarithms

Logarithms are the inverse operation of exponentiation and play a central role in the DSE Mathematics compulsory syllabus. They arise naturally when solving exponential equations, modelling growth and decay, and simplifying multiplicative structures into additive ones. This topic builds directly on the properties of exponential functions and connects to geometric sequences and series) in applications involving compound growth.

Definition of Logarithms

Logarithmic Notation

For $a \gt{} 0$, $a \neq 1$, and $x \gt{} 0$, the logarithmic statement

$$\log_a x = y$$

is defined to be equivalent to the exponential statement

$$a^y = x.$$

In words: "$\log_a x$" is the power to which $a$ must be raised to obtain $x$. The number $a$ is called the base of the logarithm, and $x$ is the argument.

The two conditions on the base arise because:

• If $a = 1$, then $1^y = 1$ for all $y$, so no unique logarithm exists.
• If $a \leq 0$, then $a^y$ is not defined for all real $y$ (e.g. $(-2)^{1/2}$ is not real).

The condition $x \gt{} 0$ follows from the fact that $a^y \gt{} 0$ for all real $y$ when $a \gt{} 0$, so the logarithm is only defined for positive arguments.

Special Cases

Several special values follow immediately from the definition:

$$\log_a 1 = 0 \quad \mathrm{since } a^0 = 1$$ $$\log_a a = 1 \quad \mathrm{since } a^1 = a$$ $$\log_a a^k = k \quad \mathrm{since } a^k = a^k$$

Additionally, for any base $a \gt{} 0$ with $a \neq 1$:

$$a^{\log_a x} = x \quad (x \gt{} 0)$$ $$\log_a(a^y) = y \quad (y \in \mathbb{'\{'}R{'\}'})$$

These two identities express the fact that the logarithmic and exponential functions are inverses of each other.

Examples

• $\log_2 8 = 3$ since $2^3 = 8$.
• $\log_3 81 = 4$ since $3^4 = 81$.
• $\log_5 1 = 0$ since $5^0 = 1$.
• $\log_7 7 = 1$ since $7^1 = 7$.
• $\log_{10} 1000 = 3$ since $10^3 = 1000$.
• $\log_2 \frac{1}{8} = -3$ since $2^{-3} = \frac{1}{8}$.

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Laws of Logarithms

The following laws are essential for manipulating logarithmic expressions. In all cases, $a \gt{} 0$, $a \neq 1$, and the arguments of all logarithms are positive.

Product Rule

$$\log_a(xy) = \log_a x + \log_a y$$

Derivation. Let $m = \log_a x$ and $n = \log_a y$, so that $a^m = x$ and $a^n = y$. Then:

$$xy = a^m \cdot a^n = a^{m+n}$$

By the definition of logarithms, $\log_a(xy) = m + n = \log_a x + \log_a y$.

Quotient Rule

$$\log_a \frac{x}{y} = \log_a x - \log_a y$$

Derivation. Similarly, let $m = \log_a x$ and $n = \log_a y$:

$$\frac{x}{y} = \frac{a^m}{a^n} = a^{m-n}$$

Therefore $\log_a \dfrac{x}{y} = m - n = \log_a x - \log_a y$.

Power Rule

$$\log_a(x^n) = n \log_a x$$

Derivation. Let $m = \log_a x$, so that $a^m = x$. Then:

$$x^n = (a^m)^n = a^{mn}$$

Therefore $\log_a(x^n) = mn = n \log_a x$.

Change of Base Formula

For any positive $c \neq 1$:

$$\log_a b = \frac{\log_c b}{\log_c a}$$

Derivation. Let $y = \log_a b$, so that $a^y = b$. Taking logarithms base $c$ of both sides:

$$\log_c(a^y) = \log_c b$$

By the power rule, $y \log_c a = \log_c b$, hence:

$$y = \frac{\log_c b}{\log_c a}$$

A commonly used special case is the change to base 10:

$$\log_a b = \frac{\log_{10} b}{\log_{10} a}$$

Examples

• $\log_2 6 + \log_2 3 = \log_2(6 \times 3) = \log_2 18$.
• $\log_3 54 - \log_3 2 = \log_3 \dfrac{54}{2} = \log_3 27 = 3$.
• $3\log_5 2 = \log_5(2^3) = \log_5 8$.
• $\log_4 9 = \dfrac{\log_{10} 9}{\log_{10} 4} = \dfrac{2\log_{10} 3}{2\log_{10} 2} = \dfrac{\log_{10} 3}{\log_{10} 2}$.

A Note on Invalid Manipulations

A common error is to distribute a logarithm over addition or subtraction:

$$\log_a(x + y) \neq \log_a x + \log_a y$$ $$\log_a(x - y) \neq \log_a x - \log_a y$$

The product and quotient rules only apply to products and quotients inside the logarithm, not to sums or differences.

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Common and Natural Logarithms

Common Logarithm (Base 10)

The common logarithm of $x$, written $\log_{10} x$ (or simply $\log x$ in many DSE contexts), is the logarithm with base 10. It is the default logarithm on most calculators and is widely used in scientific measurement scales.

Natural Logarithm (Base $e$)

The natural logarithm of $x$, written $\ln x = \log_e x$, uses the base $e \approx 2.71828$. The number $e$ is defined as:

$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$$

The natural logarithm arises naturally in calculus and in continuous growth models. Its importance stems from the fact that the derivative of $\ln x$ is $\frac{1}{x}$, making it the unique logarithm with this property.

Relationship to Exponential Functions

The exponential and logarithmic functions are inverse functions of each other. This means their graphs are reflections of each other across the line $y = x$.

Property	$y = a^x$ (exponential)	$y = \log_a x$ (logarithmic)
Domain	$\mathbb{'\{'}R{'\}'}$	$(0, \infty)$
Range	$(0, \infty)$	$\mathbb{'\{'}R{'\}'}$
$x$-intercept	$(0, 1)$ (since $a^0 = 1$)	$(1, 0)$ (since $\log_a 1 = 0$)
Asymptote	Horizontal: $y = 0$	Vertical: $x = 0$
Monotonicity	Strictly increasing when $a \gt{} 1$	Strictly increasing when $a \gt{} 1$

When $0 \lt{} a \lt{} 1$, both functions are strictly decreasing.

Examples

• $\log 1000 = 3$ since $10^3 = 1000$.
• $\log 0.01 = -2$ since $10^{-2} = 0.01$.
• $\ln e^5 = 5$ by the special case $\log_a a^k = k$.
• $\ln 1 = 0$ since $e^0 = 1$.
• $e^{\ln 7} = 7$ by the inverse property.

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Solving Logarithmic Equations

General Strategy

To solve equations involving logarithms:

1. Combine logarithmic terms using the laws of logarithms where possible.
2. Rewrite the equation in the form $\log_a(\mathrm{expression}) = k$.
3. Convert to exponential form: $\mathrm{expression} = a^k$.
4. Solve the resulting algebraic equation.
5. Check all solutions against the domain restriction: every argument of a logarithm must be positive.

Domain Restrictions

Before solving, always identify the domain. For an equation containing $\log_a f(x)$, we require $f(x) \gt{} 0$. Solutions that violate this condition are extraneous and must be discarded.

Common Mistakes

• Forgetting to check that arguments are positive.
• Applying logarithm laws to sums (e.g., writing $\log(x + 3)$ as $\log x + \log 3$).
• Dropping the base or confusing bases during a multi-step solution.
• Squaring both sides of an equation and introducing extraneous solutions.

Examples

Example 1. Solve $\log_3(x + 2) + \log_3(x - 6) = 3$.

• Domain: $x + 2 \gt{} 0$ and $x - 6 \gt{} 0$, so $x \gt{} 6$.
• Combine: $\log_3[(x+2)(x-6)] = 3$.
• Convert: $(x+2)(x-6) = 3^3 = 27$.
• Expand: $x^2 - 4x - 12 = 27$, so $x^2 - 4x - 39 = 0$.
• Quadratic formula: $x = \dfrac{4 \pm \sqrt{16 + 156}}{2} = \dfrac{4 \pm \sqrt{172}}{2} = 2 \pm \sqrt{43}$.
• Since $x \gt{} 6$, only $x = 2 + \sqrt{43}$ is accepted.

Example 2. Solve $\log_2(x) - \log_2(x - 2) = 3$.

• Domain: $x \gt{} 0$ and $x \gt{} 2$, so $x \gt{} 2$.
• Quotient rule: $\log_2 \dfrac{x}{x-2} = 3$.
• Convert: $\dfrac{x}{x-2} = 2^3 = 8$.
• Solve: $x = 8x - 16$, so $7x = 16$, hence $x = \dfrac{16}{7}$.
• Check: $\dfrac{16}{7} \gt{} 2$ is satisfied.

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Solving Exponential Equations

Taking Logarithms of Both Sides

When an equation involves terms of the form $a^{f(x)}$, taking logarithms of both sides can convert the equation from exponential to algebraic form. The choice of base is flexible; in the DSE, base 10 is most common since calculators provide direct access to $\log_{10}$.

Using the Change of Base Formula

When the equation involves different bases, rewrite all terms using the same base or apply the change of base formula to bring all logarithms to a common base.

General Strategy

1. Isolate the exponential term if possible.
2. Take logarithms of both sides.
3. Use the power rule to bring down exponents.
4. Solve the resulting linear (or polynomial) equation.

Examples

Example 1. Solve $3^{2x+1} = 7$.

• Take $\log$ of both sides: $\log(3^{2x+1}) = \log 7$.
• Power rule: $(2x+1)\log 3 = \log 7$.
• Solve: $2x + 1 = \dfrac{\log 7}{\log 3}$, so $x = \dfrac{1}{2}\left(\dfrac{\log 7}{\log 3} - 1\right)$.

Example 2. Solve $5^{x} = 2^{x+3}$.

• Take $\log$ of both sides: $\log(5^x) = \log(2^{x+3})$.
• Power rule: $x\log 5 = (x+3)\log 2$.
• Expand: $x\log 5 = x\log 2 + 3\log 2$.
• Collect: $x(\log 5 - \log 2) = 3\log 2$.
• Solve: $x = \dfrac{3\log 2}{\log 5 - \log 2} = \dfrac{3\log 2}{\log \frac{5}{2}}$.

Example 3. Solve $4^{x} - 2^{x+1} - 3 = 0$.

• Note $4^x = (2^2)^x = 2^{2x}$. Let $u = 2^x$ ($u \gt{} 0$).
• Substitute: $u^2 - 2u - 3 = 0$.
• Factor: $(u - 3)(u + 1) = 0$, so $u = 3$ or $u = -1$.
• Since $u \gt{} 0$, only $u = 3$, giving $2^x = 3$.
• Take $\log$: $x\log 2 = \log 3$, so $x = \dfrac{\log 3}{\log 2}$.

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Applications

pH Scale

The pH of a solution is defined as:

$$\mathrm{pH} = -\log_{10}[\mathrm{H}^+]$$

where $[\mathrm{H}^+]$ is the concentration of hydrogen ions (in mol/L). A lower pH means a higher concentration of hydrogen ions and therefore a more acidic solution.

• A neutral solution (pure water) has $[\mathrm{H}^+] = 10^{-7}$ mol/L, giving $\mathrm{pH} = 7$.
• Acidic solutions have $\mathrm{pH} \lt{} 7$; alkaline solutions have $\mathrm{pH} \gt{} 7$.
• The scale is logarithmic: a decrease of 1 in pH corresponds to a tenfold increase in $[\mathrm{H}^+]$.

Example

A solution has $[\mathrm{H}^+] = 2.5 \times 10^{-4}$ mol/L. Find its pH.

$$\mathrm{pH} = -\log(2.5 \times 10^{-4}) = -\left(\log 2.5 + \log 10^{-4}\right) = -(\log 2.5 - 4) = 4 - \log 2.5 \approx 3.60$$

Richter Scale

The Richter magnitude $M$ of an earthquake is defined as:

$$M = \log_{10} \frac{I}{I_0}$$

where $I$ is the amplitude of seismic waves and $I_0$ is a reference amplitude (the amplitude of a "standard" earthquake).

Because the scale is logarithmic base 10, an earthquake of magnitude 6 is ten times more powerful than one of magnitude 5, and one hundred times more powerful than one of magnitude 4.

Example

An earthquake has amplitude $5000$ times the reference. Its magnitude is:

$$M = \log_{10} 5000 = \log_{10}(5 \times 10^3) = 3 + \log_{10} 5 \approx 3.70$$

Compound Interest

The compound interest formula is closely related to logarithms and geometric sequences). If a principal $P$ is invested at an annual rate $r\%$ compounded $n$ times per year for $t$ years, the accumulated amount $A$ is:

$$A = P\left(1 + \frac{r}{100n}\right)^{nt}$$

To find the time $t$ required to reach a target amount $A$, take logarithms of both sides:

$$\begin{aligned} A &= P\left(1 + \frac{r}{100n}\right)^{nt} \\ \log A &= \log P + nt \cdot \log\left(1 + \frac{r}{100n}\right) \\ t &= \frac{\log A - \log P}{n \cdot \log\left(1 + \frac{r}{100n}\right)} \end{aligned}$$

Example

\10,000$is invested at$5%$ per annum, compounded annually. How long does it take for the investment to double?

• Set $A = 2P = 20,000$, $P = 10,000$, $r = 5$, $n = 1$.
• $20000 = 10000(1.05)^t$, so $2 = (1.05)^t$.
• Take $\log$: $\log 2 = t \log 1.05$.
• $t = \dfrac{\log 2}{\log 1.05} \approx 14.2$ years.

Exponential Growth and Decay

Many natural processes follow exponential models:

• Growth: $N(t) = N_0 \cdot a^t$ where $a \gt{} 1$ (e.g., population growth, bacterial reproduction).
• Decay: $N(t) = N_0 \cdot a^t$ where $0 \lt{} a \lt{} 1$ (e.g., radioactive decay, cooling).

The half-life $T$ of a decaying quantity is the time for the quantity to reduce to half its initial value. For $N(t) = N_0 \cdot a^t$:

$$\frac{1}{2} = a^T \implies T = \frac{\log(1/2)}{\log a} = -\frac{\log 2}{\log a}$$

For continuous decay with rate $k$: $N(t) = N_0 e^{-kt}$, and the half-life is:

$$T = \frac{\ln 2}{k}$$

Example

A radioactive substance decays such that its mass after $t$ years is given by $M(t) = 500 \cdot (0.92)^t$ grams. Find the half-life.

• Set $M(T) = 250$: $250 = 500 \cdot (0.92)^T$.
• $0.5 = (0.92)^T$.
• $T = \dfrac{\log 0.5}{\log 0.92} = \dfrac{-0.3010}{-0.0362} \approx 8.31$ years.

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Logarithmic and Exponential Inequalities

Inequalities involving logarithms require careful attention to the behaviour of the logarithmic function, which depends on whether the base is greater than or less than 1.

Case 1: Base $a \gt{} 1$

When $a > 1$, $\log_a x$ is strictly increasing, so the inequality sign is preserved:

$$\log_a x \gt{} \log_a y \iff x \gt{} y$$

Case 2: Base $0 \lt{} a \lt{} 1$

When $0 < a < 1$, $\log_a x$ is strictly decreasing, so the inequality sign is reversed:

$$\log_a x \gt{} \log_a y \iff x \lt{} y$$

Examples

Example 1. Solve $\log_2(3x - 1) < 4$.

• Since the base $2 > 1$, the inequality sign is preserved: $3x - 1 < 2^4 = 16$.
• $3x < 17$, so $x < \dfrac{17}{3}$.
• Domain: $3x - 1 > 0 \implies x > \dfrac{1}{3}$.
• Solution: $\dfrac{1}{3} < x < \dfrac{17}{3}$.

Example 2. Solve $\log_{1/2}(x + 3) \geq 1$.

• Since the base $\frac{1}{2} < 1$, the inequality sign is reversed: $x + 3 \leq \left(\frac{1}{2}\right)^1 = \frac{1}{2}$.
• $x \leq -\dfrac{5}{2}$.
• Domain: $x + 3 > 0 \implies x > -3$.
• Solution: $-3 < x \leq -\dfrac{5}{2}$.

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Graphical Properties of $y = \log_a x$

The graph of $y = \log_a x$ has the following characteristics:

• Passes through the point $(1, 0)$ since $\log_a 1 = 0$.
• Passes through the point $(a, 1)$ since $\log_a a = 1$.
• Has a vertical asymptote at $x = 0$.
• When $a > 1$, the function is strictly increasing and concave down.
• When $0 < a < 1$, the function is strictly decreasing and concave up.
• The function is defined only for $x > 0$.

Logarithmic Functions

Adjust the base $a$ to see how the shape of the logarithmic curve changes between $a > 1$ and $0 < a < 1$.

Transformations

The standard transformations apply, following the same principles as for other functions):

Transformation	Effect
$y = \log_a x + c$	Vertical shift by $c$ units
$y = \log_a(x - h)$	Horizontal shift by $h$ units
$y = -\log_a x$	Reflection in the $x$-axis
$y = \log_a(-x)$	Reflection in the $y$-axis (domain becomes $x < 0$)
$y = k\log_a x$	Vertical stretch by factor $k$

Example

Sketch $y = \log_2(x - 3) + 1$.

• Start from $y = \log_2 x$.
• Shift right by 3 units: the vertical asymptote moves from $x = 0$ to $x = 3$.
• Shift up by 1 unit: the $x$-intercept moves from $(1, 0)$ to $(1+3, 0+1) = (4, 1)$.
• The new $x$-intercept satisfies $\log_2(x - 3) + 1 = 0$, i.e. $\log_2(x-3) = -1$, giving $x - 3 = \frac{1}{2}$, so $x = 3.5$.
• Domain: $x > 3$.

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Wrap-up Questions

Question 1. Solve $\log_3(x^2 - 4) - \log_3(x + 2) = 1$.

Answer

• Domain: $x^2 - 4 > 0$ and $x + 2 > 0$, so $x > 2$.
• Quotient rule: $\log_3 \dfrac{x^2 - 4}{x + 2} = 1$.
• Factor numerator: $\dfrac{(x-2)(x+2)}{x+2} = x - 2$ (valid since $x \neq -2$).
• Convert: $x - 2 = 3^1 = 3$, so $x = 5$.
• Check: $5 > 2$ is satisfied. Solution: $x = 5$.

Question 2. Solve $2^{3x-1} = 5^{x+1}$. Give your answer in terms of logarithms.

Answer

• Take $\log$ of both sides: $\log(2^{3x-1}) = \log(5^{x+1})$.
• Power rule: $(3x - 1)\log 2 = (x + 1)\log 5$.
• Expand: $3x\log 2 - \log 2 = x\log 5 + \log 5$.
• Collect $x$ terms: $x(3\log 2 - \log 5) = \log 5 + \log 2$.
• $x = \dfrac{\log 5 + \log 2}{3\log 2 - \log 5} = \dfrac{\log 10}{3\log 2 - \log 5} = \dfrac{1}{3\log 2 - \log 5}$.

Question 3. Simplify $\dfrac{\log_8 27}{\log_8 9}$ without using a calculator.

Answer

• Let the expression equal $y$. By the change of base formula, $y = \log_9 27$.
• Write bases and argument as powers of 3: $y = \dfrac{\log_3 27}{\log_3 9} = \dfrac{3}{2}$.
• Alternatively: $\log_8 27 = \dfrac{\log 27}{\log 8} = \dfrac{3\log 3}{3\log 2}$ and $\log_8 9 = \dfrac{2\log 3}{3\log 2}$, so the ratio is $\dfrac{3\log 3 / (3\log 2)}{2\log 3 / (3\log 2)} = \dfrac{3}{2}$.

Question 4. The population of a bacteria culture grows exponentially. At 12:00, the population is $10,000$. At 14:00, the population is $40,000$. Find the population at 17:00.

Answer

• Model: $P(t) = P_0 \cdot a^t$ where $t$ is in hours from 12:00.
• At $t = 0$: $P_0 = 10,000$.
• At $t = 2$: $40,000 = 10,000 \cdot a^2$, so $a^2 = 4$, giving $a = 2$.
• At $t = 5$ (17:00): $P(5) = 10,000 \cdot 2^5 = 320,000$.

Question 5. Solve $9^x - 6 \cdot 3^x - 27 = 0$.

Answer

• Note $9^x = (3^2)^x = 3^{2x} = (3^x)^2$. Let $u = 3^x$ ($u > 0$).
• Substitute: $u^2 - 6u - 27 = 0$.
• Factor: $(u - 9)(u + 3) = 0$, so $u = 9$ or $u = -3$.
• Since $u > 0$, only $u = 9$, giving $3^x = 9 = 3^2$, so $x = 2$.

Question 6. A substance has a half-life of 8 years. How long does it take for 90% of the substance to decay?

Answer

• Model: $M(t) = M_0 \cdot a^t$ where $a = 2^{-1/8}$ (since $M(8) = \frac{1}{2}M_0$).
• 90% decay means $M(t) = 0.1M_0$.
• $0.1 = a^t = \left(2^{-1/8}\right)^t = 2^{-t/8}$.
• Take $\log$: $\log 0.1 = -\dfrac{t}{8} \log 2$.
• $t = \dfrac{-8 \log 0.1}{\log 2} = \dfrac{8}{\log 2} \approx 26.6$ years.

Question 7. If $\log_2 3 = a$ and $\log_2 5 = b$, express $\log_2 7.5$ in terms of $a$ and $b$.

Answer

• $7.5 = \dfrac{15}{2} = \dfrac{3 \times 5}{2}$.
• $\log_2 7.5 = \log_2 3 + \log_2 5 - \log_2 2 = a + b - 1$.

Question 8. Solve the inequality $\log_{0.5}(2x + 1) > \log_{0.5}(x + 4)$.

Answer

• Domain: $2x + 1 > 0 \implies x > -\dfrac{1}{2}$, and $x + 4 > 0 \implies x > -4$. Combined: $x > -\dfrac{1}{2}$.
• Since the base $0.5 < 1$, the logarithmic function is strictly decreasing, so the inequality sign reverses: $2x + 1 < x + 4$.
• $x < 3$.
• Combined with the domain: $-\dfrac{1}{2} < x < 3$.

Question 9. Express $2\log x - \log(x^2 - 4) + \log(x + 2)$ as a single logarithm, stating any restrictions on $x$.

Answer

• Restrictions: $x > 0$, $x^2 - 4 > 0 \implies x > 2$ or $x < -2$, and $x + 2 > 0 \implies x > -2$. Combined: $x > 2$.
• Combine: $2\log x + \log(x + 2) - \log(x^2 - 4) = \log(x^2) + \log(x+2) - \log(x^2 - 4)$.
• $= \log\dfrac{x^2(x+2)}{x^2 - 4} = \log\dfrac{x^2(x+2)}{(x-2)(x+2)} = \log\dfrac{x^2}{x-2}$.
• Note: the factor $(x+2)$ cancels since $x + 2 \neq 0$ under the domain $x > 2$.

Question 10. An investor deposits \5,000$into an account earning$4%$interest compounded quarterly. How long (to the nearest quarter) does it take for the balance to reach$$10,000$?

Answer

• $A = P\left(1 + \dfrac{r}{100n}\right)^{nt}$ with $P = 5000$, $r = 4$, $n = 4$, $A = 10000$.
• $10000 = 5000\left(1 + \dfrac{4}{400}\right)^{4t} = 5000(1.01)^{4t}$.
• $2 = (1.01)^{4t}$.
• Take $\log$: $\log 2 = 4t \cdot \log 1.01$.
• $t = \dfrac{\log 2}{4\log 1.01} \approx \dfrac{0.3010}{4 \times 0.00432} \approx 17.42$ years.
• To the nearest quarter: approximately 17 years and 2 quarters (17.5 years).

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tip

Diagnostic Test Ready to test your understanding of Logarithms? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Logarithms with other DSE mathematics topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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DSE Exam Technique

Showing Working

For logarithm problems in DSE Paper 1:

1. Always state the domain restriction before solving (argument of logarithm must be positive).
2. When using logarithm laws, write the law explicitly before applying it.
3. When converting between exponential and logarithmic form, show the intermediate step.
4. After solving, verify each solution satisfies the domain restriction.

Significant Figures

Exact logarithmic answers are preferred. If a numerical approximation is required, use 3 significant figures. When the question asks for an answer "correct to 3 significant figures," the calculator value should be stated.

Common DSE Question Types

1. Solving logarithmic equations (combine, convert, solve, check domain).
2. Solving exponential equations (take logs, bring down exponents).
3. Change of base problems.
4. Logarithmic inequalities (consider the base: $> 1$ or $< 1$).
5. Applications (compound interest, half-life, pH, Richter scale).

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Additional Worked Examples

Worked Example 11: Logarithmic equation with different bases

Solve $\log_2 x + \log_3 x = 5$.

Solution

By the change of base formula: $\log_3 x = \dfrac{\log_2 x}{\log_2 3}$.

Let $u = \log_2 x$:

$u + \frac{u}{\log_2 3} = 5 \implies u\!\left(1 + \frac{1}{\log_2 3}\right) = 5$

$u \cdot \frac{\log_2 3 + 1}{\log_2 3} = 5 \implies u = \frac{5\log_2 3}{1 + \log_2 3}$

$\log_2 x = \frac{5\log_2 3}{\log_2 6} = \log_6 3^5 = \log_6 243$

$x = 2^{\frac{5\log_2 3}{\log_2 6}} = 243^{1/\log_2 6}$

This can also be written as $x = 6^{\log_6 243} = 243$? Let me verify: $\log_2 243 + \log_3 243 = \log_2 243 + \dfrac{\log_2 243}{\log_2 3} = \log_2 243\!\left(1 + \dfrac{1}{\log_2 3}\right)$.

$\log_2 243 = \log_2 3^5 = 5\log_2 3$.

$5\log_2 3 \cdot \dfrac{1 + \log_2 3}{\log_2 3} = 5(1 + \log_2 3)$.

Is this $= 5$? Only if $\log_2 3 = 0$, which is false. So $x = 243$ is incorrect.

The exact answer is $x = 2^{\frac{5\log_2 3}{\log_2 6}}$.

Worked Example 12: Exponential equation with three terms

Solve $4^x - 5 \cdot 2^x + 6 = 0$.

Solution

Let $u = 2^x$ ($u > 0$): $u^2 - 5u + 6 = 0 \implies (u - 2)(u - 3) = 0$.

$u = 2$ or $u = 3$.

$2^x = 2 \implies x = 1$.

$2^x = 3 \implies x = \log_2 3 = \dfrac{\log 3}{\log 2} \approx 1.58$.

Worked Example 13: Logarithmic identity proof

Prove that $\log_a b \cdot \log_b c \cdot \log_c a = 1$.

Solution

Using the change of base formula:

$\log_a b = \frac{\log b}{\log a}, \quad \log_b c = \frac{\log c}{\log b}, \quad \log_c a = \frac{\log a}{\log c}$

Multiplying:

$\frac{\log b}{\log a} \cdot \frac{\log c}{\log b} \cdot \frac{\log a}{\log c} = 1 \qed$

Worked Example 14: Solving a system of exponential equations

Solve the system $2^x \cdot 3^y = 108$ and $2^x + 3^y = 21$.

Solution

From the first equation: $2^x = \dfrac{108}{3^y}$.

Substituting into the second: $\dfrac{108}{3^y} + 3^y = 21$.

Let $u = 3^y$ ($u > 0$): $\dfrac{108}{u} + u = 21 \implies u^2 - 21u + 108 = 0 \implies (u - 9)(u - 12) = 0$.

$u = 9$ or $u = 12$.

Case 1: $3^y = 9 \implies y = 2$. Then $2^x = \dfrac{108}{9} = 12 \implies x = \log_2 12$.

Case 2: $3^y = 12 \implies y = \log_3 12$. Then $2^x = \dfrac{108}{12} = 9 \implies x = \log_2 9$.

Solutions: $(x, y) = (\log_2 12,\; 2)$ and $(\log_2 9,\; \log_3 12)$.

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DSE Exam-Style Questions

DSE Practice 1. Solve $\log_2(2x - 1) + \log_2(x + 3) = 4$.

Solution

Domain: $2x - 1 > 0 \implies x > \dfrac{1}{2}$, and $x + 3 > 0 \implies x > -3$. Combined: $x > \dfrac{1}{2}$.

$\log_2[(2x - 1)(x + 3)] = 4 \implies (2x - 1)(x + 3) = 16$.

$2x^2 + 5x - 3 = 16 \implies 2x^2 + 5x - 19 = 0$.

$x = \dfrac{-5 \pm \sqrt{25 + 152}}{4} = \dfrac{-5 \pm \sqrt{177}}{4}$.

$\dfrac{-5 + \sqrt{177}}{4} \approx \dfrac{-5 + 13.30}{4} \approx 2.08 > \dfrac{1}{2}$. Accepted.

$\dfrac{-5 - \sqrt{177}}{4} \approx -4.58 < \dfrac{1}{2}$. Rejected.

Solution: $x = \dfrac{-5 + \sqrt{177}}{4}$.

DSE Practice 2. If $\log_4 x + \log_4(x - 6) = 2$, find $x$.

Solution

Domain: $x > 0$ and $x - 6 > 0 \implies x > 6$.

$\log_4[x(x - 6)] = 2 \implies x(x - 6) = 16 \implies x^2 - 6x - 16 = 0$.

$(x - 8)(x + 2) = 0 \implies x = 8$ or $x = -2$.

Since $x > 6$: $x = 8$.

DSE Practice 3. Solve the inequality $\log_3(x - 2) \leq 2$.

Solution

Domain: $x - 2 > 0 \implies x > 2$.

Since base $3 > 1$, the inequality sign is preserved: $x - 2 \leq 3^2 = 9$.

$x \leq 11$.

Combined: $2 < x \leq 11$.

DSE Practice 4. The population of a city was 2 million in 2020 and 3 million in 2025. Assuming exponential growth, in what year will the population reach 5 million?

Solution

$P(t) = P_0 \cdot a^t$ where $t$ is years from 2020.

$P_0 = 2$, $P(5) = 3$: $2a^5 = 3 \implies a^5 = 1.5 \implies a = 1.5^{1/5}$.

For $P(t) = 5$: $2 \cdot 1.5^{t/5} = 5 \implies 1.5^{t/5} = 2.5$.

$t = 5 \cdot \dfrac{\ln 2.5}{\ln 1.5} = 5 \cdot \dfrac{0.9163}{0.4055} \approx 11.3$ years.

Year: $2020 + 11.3 \approx 2031$ (during 2031).

DSE Practice 5. Simplify $\dfrac{\log_{27} 8}{\log_{\sqrt{3}} 2}$.

Solution

$\log_{27} 8 = \frac{\log 8}{\log 27} = \frac{3\log 2}{3\log 3} = \frac{\log 2}{\log 3}$

$\log_{\sqrt{3}} 2 = \frac{\log 2}{\log \sqrt{3}} = \frac{\log 2}{\frac{1}{2}\log 3} = \frac{2\log 2}{\log 3}$

$\frac{\log_{27} 8}{\log_{\sqrt{3}} 2} = \frac{\log 2 / \log 3}{2\log 2 / \log 3} = \frac{1}{2}$

DSE Practice 6. Given that $2\log_a x - \log_a(x^2 - 1) = 1$, express $x$ in terms of $a$.

Solution

Domain: $x > 0$, $x^2 - 1 > 0 \implies x > 1$.

$\log_a\dfrac{x^2}{x^2 - 1} = 1 \implies \dfrac{x^2}{x^2 - 1} = a$.

$x^2 = a(x^2 - 1) \implies x^2 = ax^2 - a \implies x^2(a - 1) = a$.

$x^2 = \dfrac{a}{a - 1}$.

$x = \sqrt{\dfrac{a}{a - 1}}$ (positive root since $x > 1 > 0$).

This requires $\dfrac{a}{a - 1} > 0$, i.e., $a > 1$ or $a < 0$.

Also need $x > 1$: $\dfrac{a}{a - 1} > 1 \implies \dfrac{a - a + 1}{a - 1} > 0 \implies \dfrac{1}{a - 1} > 0 \implies a > 1$.

Answer: $x = \sqrt{\dfrac{a}{a - 1}}$ for $a > 1$.