Mathematics 9709/22 — May/June 2010

Walkthrough

We want to turn the equation $13^x = (2.8)^y$ into the linear form $y = kx$. The unknown is in the exponents, so logarithms are the natural tool: taking logarithms of both sides brings the powers down as multipliers.

1. Take logarithms of both sides. Any consistent base works; the common logarithm or natural logarithm both give the same value of $k$.
2. Apply the power law $\log(a^b) = b\log a$ to each side. This gives $x\log 13 = y\log 2.8$.
3. Rearrange to make $y$ the subject by dividing both sides by $\log 2.8$: $y = \frac{\log 13}{\log 2.8}x$.
4. Evaluate the constant ratio using a calculator: $\log 13 \approx 2.56495$ and $\log 2.8 \approx 1.02962$, so $k \approx 2.49$.

Key Takeaways

This question tests the fundamental logarithmic technique of solving equations where the variable appears in an exponent. The key idea is that logarithms convert a power into a coefficient, allowing us to solve for the variable. It also shows how an exponential equation can be rewritten as a linear relationship between two variables, which is central to many later topics such as transforming data to linear form.

Common Mistakes

• Forgetting to apply the power law correctly, e.g. writing $13^x = x\log 13$ on one side but not the other.
• Mixing up which logarithm goes in the numerator and which in the denominator. Since we solve for $y$, the coefficient is $\frac{\log 13}{\log 2.8}$, not $\frac{\log 2.8}{\log 13}$.
• Using different bases on the two sides, which would give an incorrect equation unless the bases are consistent.
• Rounding too early: using rounded values of $\log 13$ and $\log 2.8$ before dividing can change the final 3-significant-figure answer.

Things to Be Careful About

• The mark scheme requires the intermediate form $y\log 2.8 = x\log 13$ to be stated or implied, so show this line explicitly.
• Any base of logarithms is acceptable, but it must be the same on both sides.
• Give the final value of $k$ to 3 significant figures: $2.49$. An unsimplified expression such as $\frac{\log 13}{\log 2.8}$ is not enough for the final mark unless the numerical value is given.
• Ensure the calculator is in the correct mode and that the logarithms are evaluated accurately before dividing.