Surds, Indices and Logarithms | A-Maths Study Guide

Secondary 3 and 4 Additional Mathematics · Study guide

Simplify surds, solve exponential and logarithmic equations, sketch inverse graphs and interpret growth and decay, with exact worked methods and topic practice.

Concept → worked examples → practice32 min read · practice linked below

Keep exact values and valid inputs

Surds, powers and logarithms are different ways of describing numbers. A surd preserves an exact irrational value; an index describes repeated multiplication and its extensions; a logarithm asks which index produces a positive number. The method works only on the inputs for which the original expression exists.

Conditions to keep beside the working
Expression	Real-number condition
$\sqrt{x}$	$x\ge0$ ; the principal square root is non-negative.
$1/\sqrt{x}$	$x>0$ , since the denominator must also be nonzero.
$a^x$ , for every real $x$	Use $a>0$ . For one-to-one exponential functions also require $a\ne1$ .
$\log_a x$	$a>0,\ a\ne1,\ x>0$ .

In particular, $\sqrt{x^2}=|x|$ , not always $x$ . Keep exact fractions, surds and logarithms until the requested final rounding.

Surds: simplify, rationalise and solve

For $u,v\ge0$ , $\sqrt{uv}=\sqrt u\sqrt v$ . Remove square factors before combining like surds: $\sqrt{12}=2\sqrt3$ . Addition has no corresponding rule: $\sqrt{u+v}$ is generally not $\sqrt u+\sqrt v$ .

To rationalise a denominator, multiply numerator and denominator by the same nonzero expression. Conjugates work because $(p+q)(p-q)=p^2-q^2$ , eliminating the radical cross terms.

The larger square consists of four equal squares of area 3. Its side is twice the smaller side; a square factor of 4 therefore contributes a factor of 2 outside the root.

Example 1: simplifying and rationalising exactly

Simplify $\sqrt{72}-\sqrt8$ , then rationalise $4/(\sqrt3-1)$ .

\sqrt{72}-\sqrt8=6\sqrt2-2\sqrt2=4\sqrt2.

\frac4{\sqrt3-1}=\frac{4(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\frac{4(\sqrt3+1)}2=2\sqrt3+2.

The denominator and conjugate are both nonzero. Multiplying the final result by $\sqrt3-1$ gives 4, confirming the value.

Example 2: a radical equation and an unwanted root

Solve $\sqrt{x+5}=x-1$ .

The left side is non-negative, so $x-1\ge0$ , giving $x\ge1$ ; this also ensures $x+5\ge0$ . Squaring gives

x+5=(x-1)^2\quad\Longrightarrow\quad x^2-3x-4=(x-4)(x+1)=0.

The candidates are 4 and −1. At $x=4$ , both original sides equal 3. At $x=-1$ , the sides are 2 and −2, so it fails. The only solution is $x=4$ . Squaring loses the distinction between a number and its negative, which is why checking the original equation is essential.

With two radicals, isolate one before squaring; if another radical remains, isolate it and square again. Record the domains and check all candidates at the end. If rational numbers $p,q$ satisfy $p+q\sqrt d=0$ with irrational $\sqrt d$ , then $p=q=0$ : otherwise $\sqrt d=-p/q$ would be rational. This justifies comparing rational and surd parts.

Indices: powers, roots and equations

For positive bases, the index laws are

a^ma^n=a^{m+n},\quad \frac{a^m}{a^n}=a^{m-n},\quad (a^m)^n=a^{mn},\quad (ab)^n=a^nb^n,\quad \left(\frac ab\right)^n=\frac{a^n}{b^n}.

Also $a^0=1$ , $a^{-n}=1/a^n$ , and $a^{p/q}=\sqrt[q]{a^p}$ for integers $p,q$ with $q>0$ . Positive bases make these rules safe for arbitrary real powers. Negative bases may allow particular integer or odd-root powers, but do not support all real exponents.

Example 3: fractional indices and a common base

Evaluate $16^{-3/4}$ , and solve $8^{x-1}=4^{x+1}$ .

16^{-3/4}=\frac1{(\sqrt[4]{16})^3}=\frac18.

Write both equation sides as powers of 2:

2^{3x-3}=2^{2x+2}\quad\Longrightarrow\quad 3x-3=2x+2\quad\Longrightarrow\quad x=5.

Equal powers have equal exponents because $2^x$ is strictly increasing. At $x=5$ , both sides are $2^{12}=4096$ .

Logarithms: the index is the answer

The definition $\log_a x=y\iff a^y=x$ explains $\log_a1=0$ , $\log_aa=1$ , $a^{\log_a x}=x$ for $x>0$ , and $\log_a(a^t)=t$ . Write $\lg x$ for base 10 and $\ln x$ for base $e$ , where $e\approx2.71828$ .

For positive $u,v$ and a valid base $a$ ,

\log_a(uv)=\log_a u+\log_a v,\quad \log_a(u/v)=\log_a u-\log_a v,\quad \log_a(u^k)=k\log_a u.

These follow from the corresponding index laws. For example, if $u=a^p$ and $v=a^q$ , then $uv=a^{p+q}$ , so its logarithm is $p+q$ . There is no law turning $\log_a(u+v)$ into a sum of logarithms.

To change to any valid base $b$ , use $\log_a x=\log_b x/\log_b a$ . Taking $\log_b$ of $a^y=x$ gives $y\log_ba=\log_bx$ ; division is valid since $a\ne1$ .

Example 4: laws and change of base

Evaluate $\log_2 12-\log_2 3+\log_4 8$ exactly.

\log_2(12/3)+\frac{\log_2 8}{\log_2 4}=2+\frac32=\frac72.

All arguments are positive. Combining first avoids rounding the two irrational logarithms separately.

Example 5: combine logarithms, then check the domain

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

Both arguments must be positive: $x>3$ . Within that domain,

\log_2[(x-1)(x-3)]=3\quad\Longrightarrow\quad (x-1)(x-3)=8.

x^2-4x-5=(x-5)(x+1)=0.

Only $x=5$ is in the original domain. It gives $\log_2 4+\log_2 2=2+1=3$ . The candidate $x=-1$ makes the product positive but each original logarithm undefined, so a combined product alone is not a sufficient domain check.

Example 6: logarithms with different bases

Solve $\log_2 x-3\log_x 2=2$ .

The variable base requires $x>0,\ x\ne1$ . Put $u=\log_2x$ , so $u\ne0$ and $\log_x2=1/u$ . Then

u-\frac3u=2\quad\Longrightarrow\quad u^2-2u-3=(u-3)(u+1)=0.

Thus $u=3$ or $-1$ , giving $x=8$ or $x=1/2$ . Both satisfy the domain and original equation. A negative logarithm is allowed: $\log_2(1/2)=-1$ ; a negative logarithm argument is not.

Exponential equations and substitutions

When a common base is inconvenient, take logarithms after isolating the exponential. When $a^{2x}$ and $a^x$ both appear, use $u=a^x>0$ , because $a^{2x}=u^2$ .

Example 7: an exponential equation without a common base

Solve $5^{2x-1}=7$ .

(2x-1)\ln5=\ln7\quad\Longrightarrow\quad x=\frac12\left(1+\frac{\ln7}{\ln5}\right)\approx1.105.

The exact answer is valid for every real $x$ in the working. Substituting its exact value makes the exponent $\ln7/\ln5$ , whose power of 5 is 7. The displayed decimal is rounded to 3 decimal places.

Example 8: a quadratic in an exponential

Solve $9^x-2(3^x)-3=0$ .

Let $u=3^x>0$ . Then $u^2-2u-3=(u-3)(u+1)=0$ . Reject $u=-1$ , since $3^x$ is always positive. The remaining equation $3^x=3$ gives $x=1$ . Direct substitution gives $9-6-3=0$ .

Graphs: growth, decay and inverse pairs

Basic exponential and logarithmic graphs
Function	Domain and range	Key features
$y=a^x$	Domain $\mathbb R$ ; range $y>0$	Passes through $(0,1)$ ; horizontal asymptote $y=0$
$y=\log_ax$	Domain $x>0$ ; range $\mathbb R$	Passes through $(1,0)$ ; vertical asymptote $x=0$

Both increase when $a>1$ , and both decrease when $0<a<1$ . Their points have interchanged coordinates, so their graphs are reflections in $y=x$ . For nonzero $k$ , $y=ka^x+c$ has asymptote $y=c$ and lies above it if $k>0$ , below it if $k<0$ . For $y=k\ln(ax+b)$ with $k\ne0$ , require $ax+b>0$ ; when $a\ne0$ the boundary $x=-b/a$ is a vertical asymptote.

Example 9: a shifted logarithm and a graphical equation

Describe $y=2\ln(x-1)$ , and state the line needed to solve $2\ln(x-1)=x-3$ graphically.

The domain is $x>1$ . The graph has vertical asymptote $x=1$ , crosses the x-axis at $x=2$ , has no y-intercept, and has range $\mathbb R$ . It increases throughout its domain. Draw $y=x-3$ on the same axes; intersection x-coordinates solve the equation.

There is a crossing between 1 and 2 because the logarithm tends to $-\infty$ near 1, but at 2 its value 0 is above the line’s value −1. A further crossing occurs between 6 and 7: $2\ln5>3$ , while $2\ln6<4$ . The increasing logarithm eventually grows more slowly than the line. Graphical readings are estimates; algebraic or numerical checks should use the original equation.

Explore an exponential and its logarithm

Move the exponent $t$ . The point $(t,a^t)$ on the exponential becomes $(a^t,t)$ on its logarithm. Changing the base to a number below 1 reverses both graphs’ direction.

Base a Exponent t:1

The plotted window is −4 to 8 on each axis. A point outside the window is still described below.

a = 2, t = 1. 2¹ = 2, so log₂ 2 = 1. Exponential point (1, 2); logarithm point (2, 1). Both graphs increase.

Models and logarithmic scales

In $Q=Q_0e^{kt}$ , $Q_0>0$ is the initial amount. Positive $k$ gives growth and negative $k$ gives decay. Equal time intervals have the same multiplicative change. In $Q=Q_0b^t$ , the factor per unit time is $b$ ; for example, 5% annual decay uses $b=0.95$ . Apply the stated time domain and interpret units.

Example 10: fit a decay model and find a time

An amount follows $Q=200e^{-kt}$ for $t\ge0$ hours and $k>0$ . After 3 hours, $Q=100$ . Find $k$ and when $Q=25$ .

100=200e^{-3k}\quad\Longrightarrow\quad -3k=\ln(1/2)\quad\Longrightarrow\quad k=\frac{\ln2}3.

For 25 units, $e^{-kt}=1/8=2^{-3}$ , so $kt=3\ln2$ . Hence $t=9$ hours. The amount halves every 3 hours: $200\to100\to50\to25$ . It approaches zero but never reaches zero at a finite time.

Example 11: a logarithmic scale measures ratios

A scale is defined by $L=10\lg(I/I_0)$ , with $I,I_0>0$ . Two readings differ by 20 units. How many times larger is the higher intensity?

L_2-L_1=10\lg\left(\frac{I_2}{I_1}\right)=20\quad\Longrightarrow\quad \frac{I_2}{I_1}=10^2=100.

The higher intensity is 100 times the lower one. A fixed increase in the reading corresponds to multiplication of intensity, not addition. The common reference $I_0$ cancels.

Common mistakes and useful checks

Adding unlike surds. Simplify square factors first; $\sqrt2+\sqrt3$ cannot be collapsed to one square root.
Dropping the domain after squaring or combining logs. Check the original radicals and each original logarithm separately.
Confusing a negative exponent with a negative value. $2^{-3}=1/8>0$ .
Treating a sum inside a logarithm as a product. The laws apply to multiplication, division and powers.
Rejecting a negative logarithm. Its argument may be between 0 and 1; this is valid.
Forgetting a variable base cannot equal 1. Record that exclusion before changing bases.
Putting an asymptote on the wrong axis. Exponentials have a horizontal asymptote; logarithms have a vertical one.
Confusing continuous and discrete rates. In $e^{kt}$ , the multiplier after one time unit is $e^k$ , not $1+k$ .

Practise this chapter

This guide supports both Secondary 3 learning and Secondary 4 revision. Choose the question bank for your level: Secondary 3: surds, indices and logarithms or Secondary 4: surds, indices and logarithms. Attempt a question before reading its worked solution.