Solving polynomial, exponential, logarithmic and circular equations: VCE Math Methods Unit 3

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

Examples in context

Example 1. Doubling time from an exponential equation. A savings balance follows $A = 3000e^{0.05t}$. To find when it doubles, solve $e^{0.05t} = 2$, so $0.05t = \ln 2$ and $t = \frac{\ln 2}{0.05} = 20\ln 2 \approx 13.86$ years. Taking the natural log turns the exponential equation into a linear one in $t$.

Example 2. Quadratic-in-disguise from a model. A signal power satisfies $4^x - 6\cdot 2^x + 8 = 0$. Since $4^x = (2^x)^2$, let $u = 2^x$: then $u^2 - 6u + 8 = (u - 2)(u - 4) = 0$, so $u = 2$ or $u = 4$. Back-substituting, $2^x = 2$ gives $x = 1$ and $2^x = 4$ gives $x = 2$.

Try this

Q1. Solve $5^{2x - 1} = 25$ for $x$. [2 marks]

• Cue. $25 = 5^2$, so $2x - 1 = 2$, giving $x = \frac{3}{2}$.

Q2. Solve $\log_3(x) + \log_3(x - 2) = 1$ for $x$. [3 marks]

• Cue. $\log_3(x(x-2)) = 1 \Rightarrow x^2 - 2x = 3 \Rightarrow (x - 3)(x + 1) = 0$; reject $x = -1$, so $x = 3$.

Q3. Solve $\cos(2x) = \frac{1}{2}$ for $x \in [0, 2\pi]$. [3 marks]

• Cue. $2x \in [0, 4\pi]$; $2x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$; so $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$.