Mathematics 9709/21 — October/November 2020

Walkthrough

We start with two logarithms subtracted. The key is the logarithm law $\ln A - \ln B = \ln\frac{A}{B}$. This lets us combine the left side into one logarithm: $\ln\frac{2x+1}{x-3} = 2$. Because $\ln$ and the exponential function are inverse operations, taking $e$ to the power of both sides removes the logarithm: $\frac{2x+1}{x-3} = e^2$. Then it becomes a linear equation. Multiply through by $x-3$, expand the bracket, bring all $x$ terms to one side and constants to the other, factor out $x$, and divide by $e^2 - 2$. The result is $x = \frac{3e^2+1}{e^2-2}$.

Key Takeaways

• The difference of two logarithms with the same base can be written as a single logarithm of a quotient.
• To solve a logarithmic equation, exponentiate both sides with the base of the logarithm.
• After removing logarithms, solve the remaining algebraic equation carefully, especially when the unknown appears inside brackets.

Common Mistakes

• Forgetting to combine logarithms before exponentiating.
• Incorrectly applying $\ln A - \ln B$ as $\ln(A-B)$ instead of $\ln\frac{A}{B}$.
• Making a sign error when expanding $e^2(x-3)$ or moving terms across the equation.
• Not checking that the final value makes both original logarithm arguments positive (here $x>3$).

Things to Be Careful About

• The original equation is only defined when $2x+1>0$ and $x-3>0$, i.e. $x>3$. The obtained value should satisfy this; it does because $e^2>2$ gives $x>3$.
• The denominator $e^2-2$ is positive, so no division by zero occurs.
• Keep $e$ as $e^2$; do not approximate unless asked.