# Exponentials, Logarithms, Laws and Equations

> Additional Mathematics · CIE IGCSE 0606 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u6-exponentials-logarithms-laws-and-equations/

This guide covers exponential and logarithmic functions, their graphs, laws of logarithms (including change of base), and solving equations of the form $a^x = b$, aligned exactly to the CIE IGCSE Add Maths 0606 2025-2027 syllabus.

**Prerequisites:** Basic algebra manipulation and indices rules; Understanding of inverse functions and graph sketching

## Learning objectives

- Recognise $e^x$ and $\ln x$ as inverse functions, and sketch their graphs with asymptotes
- Sketch graphs of form $y = ke^{nx} + a$ and $y = k \ln(ax + b)$ for integer coefficients
- Apply all laws of logarithms, including the change of base formula, to simplify expressions
- Solve equations of the form $a^x = b$ using logarithms for both calculator and non-calculator contexts
- Avoid common exam errors when manipulating logarithmic and exponential expressions

## Exponential Function $e^x$ and Graphs of $y = ke^{nx} + a$

**Natural exponential function $e^x$** — Function of the form $f(x) = e^x$, where $e \approx 2.718$ is an irrational constant. It has a horizontal asymptote at $y=0$, domain $x \in \mathbb{R}$, range $y>0$, and is equal to its own derivative.

*Example:* When $x=0$, $e^0 = 1$, so the base graph passes through the point (0, 1).

Graphs of the form $y = ke^{nx} + a$ (with integer $k, n, a$) are transformations of the base $e^x$ graph: $k$ is a vertical stretch, $n$ scales the x-axis (negative $n$ reflects over the y-axis), and $a$ shifts the graph vertically, moving the horizontal asymptote to $y = a$.

**Worked example:** Sketch the graph of $y = 3e^{2x} - 1$, state the coordinates of the y-intercept and the equation of the horizontal asymptote.

1. 1. Identify the vertical shift: $a = -1$, so the horizontal asymptote is $y = -1$.
2. 2. Calculate the y-intercept by substituting $x=0$: $y = 3e^{0} - 1 = 3(1) - 1 = 2$, so the intercept is at (0, 2).
3. 3. The coefficient $n=2$ is positive, so the graph increases exponentially as $x$ increases, approaching $y=-1$ as $x \to -\infty$.

> **Exam tip:** Always label asymptotes and intercepts on graph sketches, as examiners award explicit marks for these features even if your sketch is not perfectly to scale.

## Natural Logarithm $\ln x$ and Graphs of $y = k \ln(ax + b)$

**Natural logarithm $\ln x$** — Inverse function of $e^x$, so $\ln(e^x) = x$ and $e^{\ln x} = x$ for all valid $x$. It has a vertical asymptote at $x=0$, domain $x>0$, and range $y \in \mathbb{R}$.

*Example:* When $x=1$, $\ln 1 = 0$, so the base graph passes through the point (1, 0).

Graphs of the form $y = k \ln(ax + b)$ (with integer coefficients) are transformations of the base $\ln x$ graph. The vertical asymptote occurs where the argument of the logarithm equals zero: $ax + b = 0$, so $x = -\frac{b}{a}$. The function is only defined for values of $x$ that make $ax + b > 0$.

**Worked example:** State the domain, vertical asymptote and x-intercept of $y = 2 \ln(3x - 6)$.

1. 1. Find the vertical asymptote: set $3x - 6 = 0$, so $x = 2$.
2. 2. Calculate the domain: the argument of the logarithm must be positive, so $3x - 6 > 0$ → $x > 2$.
3. 3. Find the x-intercept by setting $y=0$: $2 \ln(3x - 6) = 0$ → $\ln(3x - 6) = 0$ → $3x - 6 = e^0 = 1$ → $x = \frac{7}{3}$, so the intercept is at $(\frac{7}{3}, 0)$.

## Laws of Logarithms and Change of Base Formula

**Logarithm (general base)** — A logarithm answers the question: to what power must the base $a$ be raised to give $b$? So $\log_a b$ is the exponent $c$ for which $a^c = b$, which makes taking a logarithm the inverse of raising the base to a power. This requires $a > 0$, $a \neq 1$ and $b > 0$.

*Notation:* $\log_a b = c \iff a^c = b$

*Example:* $\log_2 8 = 3$ because $2^3 = 8$, and $\log_{10} 1000 = 3$ because $10^3 = 1000$.

- Product rule: $\log_b (xy) = \log_b x + \log_b y$
- Quotient rule: $\log_b (\frac{x}{y}) = \log_b x - \log_b y$
- Power rule: $\log_b (x^n) = n \log_b x$
- Special cases: $\log_b b = 1$, $\log_b 1 = 0$
- Change of base formula: $\log_b a = \frac{\log_c a}{\log_c b}$ for any positive base $c \neq 1$

> **mnemonic**
>
> Remember the core log laws with PQP: Product = Plus, Quotient = Minus, Power = Multiply. This avoids mixing up operations when simplifying logarithmic expressions.

**Worked example:** Simplify $2 \log_{10} 5 + \log_{10} 8 - \log_{10} 2$, giving your answer as an integer.

1. 1. Apply the power rule to the first term: $2 \log_{10} 5 = \log_{10} 5^2 = \log_{10} 25$
2. 2. Combine using product and quotient rules: $\log_{10}25 + \log_{10}8 - \log_{10}2 = \log_{10}\left(\frac{25 \times 8}{2}\right) = \log_{10} 100$
3. 3. Evaluate: $\log_{10} 100 = 2$, since $10^2 = 100$

**Worked example:** Use the change of base formula to calculate $\log_2 15$, correct to 3 significant figures.

1. 1. Apply change of base using natural logarithms: $\log_2 15 = \frac{\ln 15}{\ln 2}$
2. 2. Substitute approximate values: $\ln 15 \approx 2.708$, $\ln 2 \approx 0.6931$
3. 3. Divide: $\frac{2.708}{0.6931} \approx 3.91$ to 3 significant figures.

> **Exam tip:** For non-calculator questions, use base 10 for change of base if the values are powers of 10 to simplify evaluation and avoid arithmetic errors.

## Solving Exponential Equations of the Form $a^x = b$

To solve exponential equations where the unknown is in the exponent, take logarithms of both sides, then apply the power rule to bring the exponent down as a multiplier. You can use any valid base for the logarithm, but natural log or base 10 are most convenient for calculation.

**Worked example:** Solve $5^{3x - 2} = 20$, giving your answer correct to 2 decimal places.

1. 1. Take natural log of both sides: $\ln(5^{3x - 2}) = \ln 20$
2. 2. Apply the power rule: $(3x - 2) \ln 5 = \ln 20$
3. 3. Rearrange to isolate the x term: $3x - 2 = \frac{\ln 20}{\ln 5} \approx \frac{2.9957}{1.6094} \approx 1.861$
4. 4. Solve for x: $3x \approx 3.861$ → $x \approx 1.29$ to 2 decimal places.

Before reaching for logarithms, check whether both sides can be written as powers of the same base. If they can, you can equate the exponents directly, which is quicker and avoids any rounding: if $a^m = a^n$ (with $a > 0$, $a \neq 1$) then $m = n$.

**Worked example:** Solve $4^{x+1} = 8^x$.

1. 1. Write both sides as powers of 2, since $4 = 2^2$ and $8 = 2^3$: $\left(2^2\right)^{x+1} = \left(2^3\right)^x$, so $2^{2(x+1)} = 2^{3x}$.
2. 2. The bases are equal, so equate the exponents: $2(x + 1) = 3x$.
3. 3. Solve: $2x + 2 = 3x$, so $x = 2$. Check: $4^{3} = 64$ and $8^{2} = 64$.

**Check your understanding**

1. What is the first step to solve $7^{x+1} = 12$?

   - Take log of both sides
   - Divide both sides by 7
   - Subtract 1 from both sides
   - Take square root of both sides

   *Answer:* Take log of both sides

   *Why:* Correct! Taking log of both sides lets you bring the unknown exponent down using the log power rule, which is the only way to isolate x in this form of equation.

## Combining an Expression into a Single Logarithm

A very common 6.2 task asks you to write an expression such as $3 + 2\lg p - \lg q$ as a single logarithm. The only extra idea beyond the product, quotient and power rules is that an independent integer constant must first be turned into a logarithm. For base 10 use $n = \lg(10^n)$, so $1 = \lg 10$, $2 = \lg 100$ and $3 = \lg 1000$; for natural logs use $n = \ln(e^n)$.

**Worked example:** Write $3 + 2\lg p - \lg q$ as a single logarithm to base 10.

1. 1. Turn the constant into a base-10 logarithm: since $10^3 = 1000$, we have $3 = \lg 1000$.
2. 2. Apply the power rule to the second term: $2\lg p = \lg p^2$.
3. 3. The expression is now $\lg 1000 + \lg p^2 - \lg q$. Combine the two added terms with the product rule: $\lg 1000 + \lg p^2 = \lg(1000 p^2)$.
4. 4. Apply the quotient rule for the subtraction: $\lg(1000 p^2) - \lg q = \lg\left(\frac{1000 p^2}{q}\right)$.
5. 5. Therefore $3 + 2\lg p - \lg q = \lg\left(\frac{1000 p^2}{q}\right)$. Check with $p = q = 1$: the original gives $3 + 0 - 0 = 3$ and the answer gives $\lg 1000 = 3$, which confirms the result.

> **Exam tip:** When a stray integer appears in a 'write as a single logarithm' question, rewrite it as a logarithm of a power of the base first (e.g. $3 = \lg 1000$), then the ordinary log laws combine everything into one logarithm.

## Solving Logarithmic Equations

Many 0606 questions give an equation containing two or more logarithms to the same base. The reliable method has four stages: use the log laws to combine each side into a single logarithm, rewrite the equation in exponential form using the definition $\log_a N = c \iff N = a^c$, solve the resulting linear or quadratic equation, and finally check every candidate in the original equation and discard any that make the argument of a logarithm zero or negative.

**Worked example:** Solve $\log_3(x + 6) + \log_3(x - 2) = 2$.

1. 1. Combine the left-hand side with the product rule: $\log_3\left[(x + 6)(x - 2)\right] = 2$.
2. 2. Rewrite in exponential form using $\log_a N = c \iff N = a^c$: $(x + 6)(x - 2) = 3^2 = 9$.
3. 3. Expand and form a quadratic: $x^2 + 4x - 12 = 9$, so $x^2 + 4x - 21 = 0$.
4. 4. Factorise and solve: $(x + 7)(x - 3) = 0$, giving $x = -7$ or $x = 3$.
5. 5. Check each root in the original equation. If $x = -7$ then $x + 6 = -1$ and $x - 2 = -9$ are both negative, so the logarithms are undefined and this root is rejected. If $x = 3$ then $x + 6 = 9 > 0$ and $x - 2 = 1 > 0$, so both logarithms are defined. The only solution is $x = 3$.

> **Exam tip:** Every root you find must be substituted back into the original equation: reject any value that makes the argument of a logarithm zero or negative, because those logarithms are undefined.

## Change of Base with a Variable Base

The change of base formula $\log_b a = \frac{\log_c a}{\log_c b}$ has a useful special case: setting the new base equal to $a$ gives $\log_x a = \frac{\log_a a}{\log_a x} = \frac{1}{\log_a x}$. This reciprocal identity lets you handle equations that mix $\log_a x$ with a term whose base is the variable, such as $\log_3 x + \log_x 3$. Substituting $y = \log_a x$ turns these into a quadratic equation. Remember that any logarithm needs $x > 0$, and when $x$ is used as a base you also need $x \neq 1$.

**Worked example:** Solve $\log_3 x + \log_x 3 = \frac{10}{3}$, where $x > 0$ and $x \neq 1$.

1. 1. Rewrite the second term with the reciprocal identity: $\log_x 3 = \frac{1}{\log_3 x}$.
2. 2. Let $y = \log_3 x$ (note $y \neq 0$ because $x \neq 1$). The equation becomes $y + \frac{1}{y} = \frac{10}{3}$.
3. 3. Multiply every term by $3y$ to clear the fractions: $3y^2 + 3 = 10y$, so $3y^2 - 10y + 3 = 0$.
4. 4. Factorise: $(3y - 1)(y - 3) = 0$, giving $y = \frac{1}{3}$ or $y = 3$.
5. 5. Convert back using $x = 3^y$: $y = \frac{1}{3} \implies x = 3^{1/3} = \sqrt[3]{3}$, and $y = 3 \implies x = 3^3 = 27$.
6. 6. Both $x = \sqrt[3]{3} \approx 1.44$ and $x = 27$ satisfy $x > 0$ and $x \neq 1$, so both are valid. Check $x = 27$: $\log_3 27 + \log_{27} 3 = 3 + \frac{1}{3} = \frac{10}{3}$, as required.

> **Exam tip:** When the unknown appears as the base of a logarithm (e.g. $\log_x 3$), use the reciprocal identity $\log_x a = \frac{1}{\log_a x}$ to write everything in one base, then substitute $y = \log_a x$ to obtain a quadratic in $y$.

## Log Graphs That Have a y-Intercept

Sketch questions on $y = k\ln(ax + b)$ frequently ask for the y-intercept. This exists only when the argument $ax + b$ is positive at $x = 0$, that is when $b > 0$. Substitute $x = 0$ to get $y = k\ln b$. Curves like $y = 2\ln(3x - 6)$ have no y-intercept because the argument is negative at $x = 0$, but a curve such as $y = \ln(x + 2)$ does.

**Worked example:** Sketch $y = \ln(x + 2)$, stating the domain, the vertical asymptote, the x-intercept and the y-intercept.

1. 1. Domain: the argument must be positive, so $x + 2 > 0 \implies x > -2$.
2. 2. Vertical asymptote: the argument is zero at $x + 2 = 0 \implies x = -2$. The curve falls towards $-\infty$ as $x \to -2^+$.
3. 3. x-intercept: set $y = 0$, so $\ln(x + 2) = 0 \implies x + 2 = e^0 = 1 \implies x = -1$. The x-intercept is $(-1, 0)$.
4. 4. y-intercept: set $x = 0$, so $y = \ln(0 + 2) = \ln 2 \approx 0.69$. The y-intercept is $(0, \ln 2)$.
5. 5. Shape: an increasing curve with vertical asymptote $x = -2$, passing through $(-1, 0)$ and $(0, \ln 2)$, rising slowly for large $x$.

> **warning**
>
> Do not assume every log graph crosses the y-axis. Check the sign of the argument at $x = 0$ first: for $y = \ln(x + 2)$ the argument is $2 > 0$, so the y-intercept exists; for $y = 2\ln(3x - 6)$ it is $-6 < 0$, so there is none.

> **Exam tip:** A logarithm graph only has a y-intercept when the argument is positive at $x = 0$, i.e. when $b > 0$ in $y = k\ln(ax + b)$. Always test $x = 0$: if the argument comes out $\le 0$, state that there is no y-intercept.

## Common pitfalls

- **Wrong:** Applying log laws to sums: $\log(x + y) = \log x + \log y$
  - Why it fails: Logarithms only turn products into sums, not additions. The product rule only applies to $\log(xy)$, not $\log(x + y)$.
  - Correct: Leave $\log(x + y)$ as written, or factor the argument if possible before applying log laws.
- **Wrong:** Forgetting the domain of $\ln x$: accepting solutions where the log argument is negative or zero.
  - Why it fails: Logarithms are only defined for strictly positive arguments, so any solution that makes $ax + b ≤ 0$ is invalid and must be discarded.
  - Correct: Always check that your solutions make the argument of any logarithm in the original equation strictly positive.
- **Wrong:** Misidentifying the asymptote of $y = ke^{nx} + a$ as $x = a$ instead of $y = a$.
  - Why it fails: Exponential functions have horizontal asymptotes, while logarithmic functions have vertical asymptotes.
  - Correct: Exponential graphs have horizontal asymptotes equal to their vertical shift $a$, logarithmic graphs have vertical asymptotes where their argument equals zero.
- **Wrong:** Misapplying the inverse identity: $e^{2 \ln x} = 2x$
  - Why it fails: The identity $e^{\ln f(x)} = f(x)$ only applies when the entire exponent is a single logarithm term.
  - Correct: Apply the power rule first to combine the coefficient into the logarithm: $e^{2 \ln x} = e^{\ln x^2} = x^2$.
- **Wrong:** Using change of base incorrectly: $\log_b a = \frac{\log_c b}{\log_c a}$
  - Why it fails: The base of the original logarithm becomes the denominator in the change of base ratio.
  - Correct: Use the mnemonic 'base goes below': $\log_{base}(value) = \frac{\log(value)}{\log(base)}$.

## Cheatsheet

| Concept | Rule / Formula | Key Exam Note |
| --- | --- | --- |
| $e^x$ properties | Domain: $\mathbb{R}$, Range: $y>0$, Asymptote: $y=0$ | Passes through (0, 1), inverse of $\ln x$ |
| $\ln x$ properties | Domain: $x>0$, Range: $\mathbb{R}$, Asymptote: $x=0$ | Passes through (1, 0), inverse of $e^x$ |
| Log Product Rule | $\log_b(xy) = \log_b x + \log_b y$ | Only applies to products, not sums |
| Log Quotient Rule | $\log_b(\frac{x}{y}) = \log_b x - \log_b y$ | Only applies to quotients, not differences |
| Log Power Rule | $\log_b(x^n) = n \log_b x$ | Applies to all real powers $n$ |
| Change of Base | $\log_b a = \frac{\log_c a}{\log_c b}$ | Use base 10 or $e$ for calculations |
| Solve $a^x = b$ | $x = \frac{\ln b}{\ln a}$ | Any valid log base is accepted in exams |

## What's next

Now that you have mastered the foundational laws, graphs, and equation-solving techniques for exponential and logarithmic functions, you are ready to apply these concepts to more advanced problems in the CIE IGCSE Add Maths 0606 syllabus. Next, you will learn how to differentiate and integrate exponential and logarithmic functions, which are frequently tested in both Paper 1 (non-calculator) and Paper 2 (calculator) exams. You will also encounter these functions when solving real-world growth and decay problems, such as population growth, radioactive decay, and compound interest scenarios. Make sure to practice a mix of calculator and non-calculator questions to build confidence, and review the common pitfalls listed above to avoid losing easy marks.

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