Algebra, indices and equations: VCE Math Methods Unit 1 Year 11

What this dot point is asking

VCAA wants you to manipulate algebraic expressions involving indices and logarithms, factorise polynomial expressions, and solve linear, quadratic, polynomial, exponential and logarithmic equations. The dot point builds the algebraic fluency Unit 3 / 4 will require.

Index laws

For real $a, b$ and integer or rational $m, n$:

Logarithm laws

For $a > 0, a \neq 1$ and positive $m, n$:

Change of base.

So $\log_2 5 = \log_{10} 5 / \log_{10} 2 \approx 2.32$.

Inverse relationship. $\log_a(a^x) = x$ and $a^{\log_a x} = x$.

Factorisation techniques

Common factor

$6 x^3 - 9 x^2 = 3 x^2 (2 x - 3)$.

Grouping

$x^3 + 2 x^2 - x - 2 = x^2(x + 2) - (x + 2) = (x + 2)(x^2 - 1) = (x + 2)(x - 1)(x + 1)$.

Quadratic factorisation

$x^2 + 5x + 6 = (x + 2)(x + 3)$. Use sum-and-product (looking for two numbers that sum to the middle coefficient and multiply to the constant).

Quadratic formula

$a x^2 + b x + c = 0$ has solutions $x = (-b \pm \sqrt{b^2 - 4ac}) / (2a)$.

Difference of squares

$a^2 - b^2 = (a - b)(a + b)$.

Sum and difference of cubes

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$; $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Solving equations

Linear

Single step or simple multi-step manipulation.

Quadratic

Factor first, then use the null factor law ($AB = 0 \implies A = 0$ or $B = 0$). Or use the quadratic formula.

Polynomial

Factor where possible. Look for rational roots first; then use polynomial division or grouping.

Exponential

Bring to common base if possible, then equate exponents. Otherwise take logarithms.

Example: $5^x = 17$. Take $\log_{10}$: $x \log_{10} 5 = \log_{10} 17$, so $x = \log_{10} 17 / \log_{10} 5 \approx 1.76$.

Logarithmic. Combine logs using laws; convert to exponential form. Always check domain (logs require positive arguments).

Example: $\log_3(x + 5) = 2$. Convert: $x + 5 = 3^2 = 9$, so $x = 4$. Check: $\log_3(9) = 2$. Confirmed.

Worked example: solving an exponential equation

Solve $4^x - 3 \cdot 2^x + 2 = 0$.

Substitute $u = 2^x$. Then $4^x = (2^x)^2 = u^2$.

Equation becomes $u^2 - 3u + 2 = 0$, factoring as $(u - 1)(u - 2) = 0$.

So $u = 1$ or $u = 2$, i.e. $2^x = 1$ or $2^x = 2$, giving $x = 0$ or $x = 1$.

Worked example: simultaneous equations

$3 x + 2 y = 14$ and $x - y = 3$.

From the second: $x = y + 3$. Substitute: $3(y + 3) + 2 y = 14$, so $5 y + 9 = 14$, $y = 1$, then $x = 4$.

Examples in context

Example 1. Compound interest as an exponential equation. A managed fund grows at $6\%$ per year, so a balance of $\$8000$ follows $A = 8000 (1.06)^t$. To find when it doubles, solve $8000(1.06)^t = 16000$, i.e. $(1.06)^t = 2$. Take $\log_{10}$: $t \log_{10} 1.06 = \log_{10} 2$, so $t = \frac{\log_{10} 2}{\log_{10} 1.06} = \frac{0.30103}{0.025306} \approx 11.9$ years.

Example 2. Factorising to solve a polynomial. A design constraint gives $x^3 - 2x^2 - 5x + 6 = 0$. Testing $x = 1$: $1 - 2 - 5 + 6 = 0$, so $(x - 1)$ is a factor. Dividing gives $x^2 - x - 6 = (x - 3)(x + 2)$. Hence the solutions are $x = 1, 3, -2$.

Try this

Q1. Solve $2^{3x + 1} = 32$. [2 marks]

• Cue. Write $32 = 2^5$, so $3x + 1 = 5$, giving $x = \tfrac{4}{3}$.

Q2. A population is modelled by $P = 500 (1.04)^t$ ($t$ in years). Find, to one decimal place, when $P = 900$. [3 marks]

• Cue. $(1.04)^t = 1.8$; $t = \log_{10} 1.8 / \log_{10} 1.04 \approx 14.99$, so about $15.0$ years.

Q3. (a) Factorise $x^3 + x^2 - 4x - 4$. (b) Hence solve $x^3 + x^2 - 4x - 4 = 0$. [2+1 marks]

• Cue. (a) Group: $x^2(x + 1) - 4(x + 1) = (x + 1)(x^2 - 4) = (x + 1)(x - 2)(x + 2)$. (b) $x = -1, 2, -2$.