Year 10A Logarithms and Log Laws Worksheets & Practice

Worked examples

Year 10A logarithms and log laws: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Converting between logarithmic and exponential form

   Write each statement in the equivalent form.

   (a) Convert $\log_3 81 = 4$ to exponential form.

   (b) Convert $5^{-2} = 0.04$ to logarithmic form.

   (c) Find the value of $\log_2 32$.

   Show worked solution

   Key idea: $\log_b x = y$ means exactly the same as $b^y = x$. You can swap between the two forms freely.

   1. (a) The base is $3$, the logarithm is $4$, and the number is $81$, so the exponential form is:
      $3^4 = 81$

   2. (b) The base is $5$, the exponent is $-2$, and the result is $0.04$, so the logarithmic form is:
      $\log_5 0.04 = -2$

   3. (c) Let $\log_2 32 = y$. This means $2^y = 32$. Since $2^5 = 32$:
      $y = 5$

   Answer: (a) $3^4 = 81$, (b) $\log_5 0.04 = -2$, (c) $\log_2 32 = 5$

2. Example 2 · Simplifying expressions using log laws

   Simplify the following expression, writing your answer as a single logarithm or integer.
   $2\log_5 10 - \log_5 4 + \log_5 2$

   Show worked solution

   Key idea: Apply the power law first ($n\log_b m = \log_b m^n$), then combine using the product and quotient laws ($\log_b m + \log_b n = \log_b(mn)$ and $\log_b m - \log_b n = \log_b\!\left(\dfrac{m}{n}\right)$).

   1. Apply the power law to the first term:
      $2\log_5 10 = \log_5 10^2 = \log_5 100$

   2. The expression becomes:
      $\log_5 100 - \log_5 4 + \log_5 2$

   3. Apply the quotient law to the first two terms:
      $\log_5 \frac{100}{4} + \log_5 2 = \log_5 25 + \log_5 2$

   4. Apply the product law:
      $\log_5(25 \times 2) = \log_5 50$

   Check: $\log_5 50$ does not simplify further since $50$ is not an integer power of $5$.

   Answer: $\log_5 50$

3. Example 3 · Logarithmic scales: comparing earthquake magnitudes

   The Richter scale measures earthquake magnitude using the formula $M = \log_{10}\!\left(\dfrac{I}{I_0}\right)$, where $I$ is the intensity of the earthquake and $I_0$ is a fixed reference intensity. An earthquake in regional Victoria is recorded at magnitude $4.2$. A second earthquake near Mildura is recorded at magnitude $6.0$. How many times more intense is the Mildura earthquake than the Victorian earthquake? Give your answer to the nearest whole number.

   Show worked solution

   Key idea: Rearrange the Richter formula to express intensity, then find the ratio of the two intensities.

   1. For an earthquake of magnitude $M$, the formula $M = \log_{10}\!\left(\dfrac{I}{I_0}\right)$ rearranges to:
      $\frac{I}{I_0} = 10^M$

   2. Write the intensity of each earthquake relative to $I_0$:
      $\begin{aligned} I_{\text{Victoria}} &= I_0 \times 10^{4.2} \\ I_{\text{Mildura}} &= I_0 \times 10^{6.0} \end{aligned}$

   3. Find the ratio of the two intensities:
      $\frac{I_{\text{Mildura}}}{I_{\text{Victoria}}} = \frac{I_0 \times 10^{6.0}}{I_0 \times 10^{4.2}} = 10^{6.0 - 4.2} = 10^{1.8}$

   4. Evaluate:
      $10^{1.8} \approx 63.1$

   Answer: The Mildura earthquake is approximately $63$ times more intense.

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