Define logarithms as the inverse of exponentials, apply the laws of logarithms, sketch logarithmic graphs and solve exponential equations using logs

What this dot point is asking

VCAA wants you to use logarithms as the inverse operation to exponentials, apply the three core log laws, and solve exponential equations.

Definition

$\log_b x = y \iff b^y = x$.

Equivalently $b^{\log_b x} = x$ and $\log_b(b^y) = y$. Logarithms and exponentials are inverse functions.

Two bases dominate:

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

The laws of logarithms

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

Special values: $\log_b 1 = 0$, $\log_b b = 1$, $\log_b(b^x) = x$.

Change of base: $\log_b a = \dfrac{\log_{10} a}{\log_{10} b} = \dfrac{\ln a}{\ln b}$.

Graph of $y = \log_b x$

For $b > 1$:

• Domain: $x > 0$. Range: all real $y$.
• Vertical asymptote: $x = 0$.
• Increasing, concave down.
• x-intercept: $(1, 0)$.

Reflects $y = b^x$ in the line $y = x$.

Solving exponential equations

If $b^x = m$ where $m$ is not a power of $b$, take log of both sides:

Solving logarithmic equations

Equations involving logarithms are solved by first combining all log terms into a single logarithm using the laws, then converting to exponential form. For example, $\log_b(P) = q$ becomes $P = b^q$. A critical final step is checking each solution against the domain: the argument of every logarithm must be strictly positive, so candidate roots that make any argument zero or negative must be rejected. This domain check is the single most-marked step in VCAA logarithmic-equation questions, because the algebra often throws up an extraneous negative root.

Solving exponential equations with logarithms

When the two sides of an exponential equation cannot be written to a common base, take the logarithm of both sides and use the power law $\log_b(x^n) = n\log_b x$ to bring the unknown exponent down to a coefficient. The equation then becomes linear in the unknown and can be rearranged. For $a^{f(x)} = c$, this gives $f(x) = \dfrac{\log c}{\log a}$. Either base ($\log_{10}$ or $\ln$) works, as the change-of-base ratio is the same; choose whichever the calculator or exam expects.

In one sentence

Logarithms invert exponentials ($\log_b x = y \iff b^y = x$); the three laws ($\log_b(xy) = \log_b x + \log_b y$, $\log_b(x/y) = \log_b x - \log_b y$, $\log_b(x^n) = n \log_b x$) plus change of base solve exponential equations whose two sides cannot be reduced to a single base.

Examples in context

Example 1. Time for an investment to grow. A deposit grows by $7\%$ per year, so the balance follows $A = A_0(1.07)^t$. To find when it triples, solve $(1.07)^t = 3$. Taking logs: $t = \frac{\log 3}{\log 1.07} = \frac{0.4771}{0.02938} \approx 16.2$ years. Logs convert the unknown exponent into a quotient that the calculator evaluates directly.

Example 2. Sound intensity in decibels. Loudness is $L = 10\log_{10}\!\left(\frac{I}{I_0}\right)$ decibels, where $I_0$ is the reference intensity. If one source measures $70$ dB, then $\frac{I}{I_0} = 10^{7} = 10{,}000{,}000$. Doubling the intensity adds $10\log_{10}2 \approx 3$ dB, illustrating the log law $\log(2I) - \log I = \log 2$.

Try this

Q1. Evaluate $\log_3 81$ and $\log_2 \frac{1}{8}$. [2 marks]

• Cue. $\log_3 81 = 4$ (since $3^4 = 81$); $\log_2 \frac{1}{8} = -3$.

Q2. Solve $3^x = 20$, giving the answer to three significant figures. [2 marks]

• Cue. $x = \frac{\log 20}{\log 3} = \frac{1.301}{0.4771} \approx 2.73$.

Q3. Simplify $\log_5 50 - \log_5 2$. [2 marks]

• Cue. $\log_5(50/2) = \log_5 25 = 2$.