Year 10A Exponential and Logarithmic Equations Worksheets & Practice

Worked examples

Year 10A exponential and logarithmic equations: 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 exponential and logarithmic form

   Write each statement in the equivalent form.

   (a) $5^3 = 125$ in logarithmic form

   (b) $\log_4 64 = 3$ in exponential form

   Show worked solution

   Key idea: $a^x = y$ and $\log_a y = x$ are two ways of saying the same thing, the base stays the base, and the exponent and the result swap roles.

   1. For $5^3 = 125$: the base is $5$, the exponent is $3$, and the result is $125$, so the logarithmic form is $\log_5 125 = 3$.
   2. For $\log_4 64 = 3$: the base is $4$, the logarithm equals $3$, and the input is $64$, so the exponential form is $4^3 = 64$.

   Answer: (a) $\log_5 125 = 3$ and (b) $4^3 = 64$

2. Example 2 · Solving a logarithmic equation using log laws

   Solve for $x$, checking that your answer is valid:
   $\log_3(x + 6) + \log_3 x = 3$

   Show worked solution

   Key idea: Use the product law $\log_a m + \log_a n = \log_a(mn)$ to combine the left side, then convert to exponential form to solve.

   1. Apply the product law:
      $\log_3\bigl(x(x+6)\bigr) = 3$
   2. Convert to exponential form:
      $x(x + 6) = 3^3 = 27$
   3. Expand and rearrange:
      $x^2 + 6x - 27 = 0$
   4. Factorise:
      $(x + 9)(x - 3) = 0 \implies x = -9 \text{ or } x = 3$
   5. Validity check: both $\log_3(x+6)$ and $\log_3 x$ require their arguments to be positive, so $x > 0$. Reject $x = -9$.

   Answer: $x = 3$

3. Example 3 · Solving a real-life exponential equation using logarithms

   A town's population is currently $8\,000$ and is growing at $6\%$ per year, so after $t$ years the population is given by $P = 8000 \times 1.06^t$. After how many complete years will the population first exceed $12\,000$? Give your answer to the nearest whole year.

   Show worked solution

   Key idea: When the unknown is in the exponent, take the logarithm of both sides and use the power law $\log(a^t) = t\log a$.

   1. Set up the inequality and solve the equation $8000 \times 1.06^t = 12000$:
      $1.06^t = \frac{12000}{8000} = 1.5$
   2. Take $\log_{10}$ of both sides:
      $\log(1.06^t) = \log(1.5)$
   3. Apply the power law:
      $t\log(1.06) = \log(1.5)$
   4. Solve for $t$:
      $t = \frac{\log(1.5)}{\log(1.06)} = \frac{0.17609\ldots}{0.02531\ldots} \approx 6.96$
   5. Since the population must first exceed $12\,000$, take the next whole year: $t = 7$.

   Answer: After $7$ complete years.

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