Solution of polynomial equations of low degree with real coefficients, exponential and logarithmic equations using properties such as $a^x = e^{x\ln a}$, and circular equations using exact unit-circle values

What this dot point is asking

VCAA wants you to solve four families of equations by hand: polynomial (using the factor theorem), exponential and logarithmic (using log laws and the change-of-base identity), and circular (using exact unit-circle values). This is Paper 1 core algebra and shows up in nearly every paper.

Polynomial equations

For $P(x) = 0$ where $P$ is a low-degree polynomial:

• Quadratics. Factor, complete the square, or use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Cubics and quartics. Apply the factor theorem to find one rational root, divide it out, then factorise the resulting quadratic.

Example. Solve $x^3 - 4x^2 + x + 6 = 0$.

Trial $x = -1$: $-1 - 4 - 1 + 6 = 0$. So $(x + 1)$ is a factor.

Divide: $x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6) = (x + 1)(x - 2)(x - 3)$.

Solutions: $x = -1, 2, 3$.

Exponential equations

Standard form, single base

If both sides can be written with the same base, equate exponents.

Example. $2^{x+1} = 8$ becomes $2^{x+1} = 2^3$, so $x + 1 = 3$ and $x = 2$.

Using the change of base

If different bases appear, convert using $a^x = e^{x \ln a}$ (or take $\ln$ of both sides).

Example. $3^x = 7$. Take $\ln$: $x \ln 3 = \ln 7$, so $x = \frac{\ln 7}{\ln 3}$.

Substitution trick for quadratic-in-exponential

Equations like $e^{2x} - 5 e^x + 6 = 0$ become quadratics under $u = e^x$.

Let $u = e^x$. Then $u^2 - 5u + 6 = 0$, so $(u - 2)(u - 3) = 0$, giving $u = 2$ or $u = 3$.

Back-substitute: $e^x = 2$ gives $x = \ln 2$; $e^x = 3$ gives $x = \ln 3$.

The same pattern works for any $e^{2x} + A e^x + B = 0$.

Logarithmic equations

Combine using log laws

If the equation contains multiple log terms, combine using the product, quotient and power laws into a single log.

Exponentiate to remove the log

Once you have $\ln(\text{expression}) = c$, exponentiate both sides: expression $= e^c$. Then solve the resulting algebraic equation.

Check for spurious solutions

Log arguments must be positive. After solving, substitute back into the original equation and reject any solutions that make a log argument zero or negative.

Example. Solve $\ln(x + 2) - \ln(x - 1) = \ln(4)$.

Combine: $\ln\!\left(\frac{x + 2}{x - 1}\right) = \ln(4)$.

So $\frac{x + 2}{x - 1} = 4$, giving $x + 2 = 4(x - 1)$, i.e. $x + 2 = 4x - 4$, so $3x = 6$ and $x = 2$.

Check: $\ln(4) - \ln(1) = \ln(4) - 0 = \ln(4)$. Valid.

Circular equations

One basic period

To solve $\sin(\theta) = k$ for $\theta$ in a given interval:

1. Find the reference angle $\alpha$ from $\sin(\alpha) = |k|$.
2. Use the ASTC quadrant rule to identify which quadrants give the correct sign.
3. List one solution in each appropriate quadrant in $[0, 2\pi]$ (or $[-\pi, \pi]$ depending on convention).
4. Add or subtract multiples of the period to extend to other intervals.

Compound argument

For $\sin(b \theta - c) = k$ on $[0, 2\pi]$, let $u = b\theta - c$. The new variable $u$ ranges over $[-c, 2\pi b - c]$. Find all solutions in this extended range, then back-substitute and solve for $\theta$.

Example. Solve $\sin(2x) = \frac{\sqrt{3}}{2}$ for $x \in [0, 2\pi]$.

Let $u = 2x \in [0, 4\pi]$.

Reference angle: $\sin(\alpha) = \sqrt{3}/2$ gives $\alpha = \pi/3$. Sine is positive in quadrants 1 and 2.

Solutions for $u$ in $[0, 4\pi]$: $u = \pi/3, 2\pi/3, \pi/3 + 2\pi, 2\pi/3 + 2\pi$, i.e. $u = \pi/3, 2\pi/3, 7\pi/3, 8\pi/3$.

Divide by 2: $x = \pi/6, \pi/3, 7\pi/6, 4\pi/3$.

Trig identities for solving

The Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$ lets you convert between $\sin$ and $\cos$. The substitution $u = \sin(x)$ (or $\cos(x)$) reduces some trig equations to quadratics.

Example. Solve $2 \sin^2(x) - \sin(x) - 1 = 0$ for $x \in [0, 2\pi]$.

Let $u = \sin(x)$. Then $2 u^2 - u - 1 = 0$, factoring as $(2u + 1)(u - 1) = 0$. So $u = -1/2$ or $u = 1$.

$\sin(x) = 1$ gives $x = \pi/2$.

$\sin(x) = -1/2$ gives $x = 7\pi/6$ or $x = 11\pi/6$.

Solutions: $x = \pi/2, 7\pi/6, 11\pi/6$.

Common Paper 1 traps

Dropping solutions when squaring. Squaring both sides of an equation can introduce extraneous roots; always substitute back to check.

Wrong interval for a substituted variable. When $u = 2x$ and $x \in [0, 2\pi]$, $u$ ranges over $[0, 4\pi]$, not $[0, 2\pi]$. Doubling the argument doubles the number of solutions.

Ignoring the domain on log equations. Spurious solutions from log equations always come back to argument-positivity. Always check.

Treating $\log(a + b)$ as $\log(a) + \log(b)$. This is false. Only $\log(ab) = \log(a) + \log(b)$.

Computing decimals on Paper 1. Exact answers like $\frac{\ln 7}{\ln 3}$, $\ln 2$, or $\pi/6$ are required.

In one sentence

Solving polynomial, exponential, logarithmic and circular equations in Paper 1 relies on a small toolkit: the factor theorem for polynomials, log laws and the change-of-base identity for exponentials and logs, exact unit-circle values for trig, and the substitution trick $u = e^x$ or $u = \sin(x)$ for equations that are quadratic in disguise.