← Unit 2: Calculus

QLDMath MethodsSyllabus dot point

Topic 1: Exponential functions

Define logarithms as the inverse of exponentials, apply the laws of logarithms, and solve exponential equations using logarithms

A focused answer to the QCE Math Methods Unit 2 dot point on logarithms. States the definition $\log_b x = y \iff b^y = x$, derives the laws (product, quotient, power, change of base), and works the QCAA-style exponential equation $5^x = 28$ using logs.

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What this dot point is asking

QCAA wants you to use logarithms as the inverse operation to exponentials, apply the three core log laws, and solve exponential equations that cannot be reduced to a single common base.

Definition

The logarithm log⁑bx\log_b x is the exponent to which the base bb must be raised to give xx:

log⁑bx=yβ€…β€ŠβŸΊβ€…β€Šby=x\log_b x = y \iff b^y = x

Equivalently, blog⁑bx=xb^{\log_b x} = x and log⁑b(by)=y\log_b (b^y) = y. Logarithms and exponentials are inverse functions.

Two bases dominate:

  • Common log (log⁑10\log_{10}, written log⁑\log): scientific notation, decibels, pH.
  • Natural log (log⁑e\log_e, written ln⁑\ln): calculus, continuous growth.

The laws of logarithms

For any positive base b≠1b \neq 1 and positive x,yx, y:

DMATH_1
DMATH_2

log⁑b(xn)=nlog⁑bx(power)\log_b (x^n) = n \log_b x \quad (\text{power})

Special values:

log⁑b1=0,log⁑bb=1,log⁑b(bx)=x\log_b 1 = 0, \quad \log_b b = 1, \quad \log_b (b^x) = x

Change of base (useful for evaluating log⁑580\log_5 80 on a calculator):

log⁑ba=log⁑10alog⁑10b=ln⁑aln⁑b\log_b a = \frac{\log_{10} a}{\log_{10} b} = \frac{\ln a}{\ln b}

Solving exponential equations

If bx=mb^x = m where mm is not a power of bb, take log of both sides.

log⁑bx=log⁑mβ€…β€ŠβŸΉβ€…β€Šxlog⁑b=log⁑mβ€…β€ŠβŸΉβ€…β€Šx=log⁑mlog⁑b\log b^x = \log m \implies x \log b = \log m \implies x = \frac{\log m}{\log b}

This is the universal method when same-base manipulation fails.

Worked example

Simplify log⁑224βˆ’log⁑23+log⁑24\log_2 24 - \log_2 3 + \log_2 4.

Apply the quotient law to the first two, then the product law:

log⁑2(24/3)+log⁑24=log⁑28+log⁑24=log⁑2(8Γ—4)=log⁑232=5\log_2 (24/3) + \log_2 4 = \log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32 = 5 (since 25=322^5 = 32).

Common traps

Treating log⁑(x+y)\log(x+y) as log⁑x+log⁑y\log x + \log y. Not a log law. The product law applies to log⁑(xy)\log(xy), not log⁑(x+y)\log(x+y).

Forgetting the base of log⁑\log. In QCAA Math Methods, log⁑\log without a base usually means log⁑10\log_{10}, and ln⁑\ln means log⁑e\log_e. Be explicit when answering.

Dividing inside the log instead of subtracting outside. log⁑(20/4)=log⁑5\log(20/4) = \log 5, but it can also be expanded as log⁑20βˆ’log⁑4\log 20 - \log 4. The two are equal.

Taking log of a negative number. log⁑b\log_b is only defined for positive arguments. Equations like log⁑(xβˆ’3)=1\log(x-3) = 1 require x>3x > 3 in the domain.

How this appears in IA1 and EA

IA1. Simplify a multi-term log expression using the laws; solve an exponential equation.

EA Paper 1. Multiple choice on log values and law applications.

EA Paper 2. A continuous compound interest or population-growth context where ln⁑\ln appears, followed by solving for time.

In one sentence

The logarithm log⁑bx=y\log_b x = y is defined by by=xb^y = x, and the three laws (log⁑b(xy)=log⁑bx+log⁑by\log_b(xy) = \log_b x + \log_b y, log⁑b(x/y)=log⁑bxβˆ’log⁑by\log_b(x/y) = \log_b x - \log_b y, log⁑b(xn)=nlog⁑bx\log_b(x^n) = n \log_b x) plus change of base let you simplify expressions and solve exponential equations whose two sides cannot be reduced to a single base.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC3 marksSolve $3 \cdot 5^{x+1} = 240$. Express the answer in exact form using logarithms, then evaluate to three significant figures.
Show worked answer β†’

Isolate the exponential.

5x+1=805^{x+1} = 80

Apply log⁑10\log_{10} to both sides.

(x+1)log⁑5=log⁑80(x+1) \log 5 = \log 80

x+1=log⁑80log⁑5x + 1 = \frac{\log 80}{\log 5}

x=log⁑80log⁑5βˆ’1x = \frac{\log 80}{\log 5} - 1 (exact form)

x=1.90310.6990βˆ’1=2.7227βˆ’1=1.72x = \frac{1.9031}{0.6990} - 1 = 2.7227 - 1 = 1.72 (three sig fig).

Markers reward isolating the exponential first, applying log to both sides, the exact-form answer, and the numerical evaluation.

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