Mathematics 9709/23 — October/November 2020

Walkthrough

Start with $\ln(2x+1) - \ln(x-3) = 2$. The expression is a difference of two logarithms with the same base, so it can be combined into a single logarithm: $\ln\left(\frac{2x+1}{x-3}\right)$. This is the key logarithm law: subtracting logs corresponds to dividing the arguments.

Next, because the natural logarithm and the exponential function are inverses, applying $e$ to both sides removes the logarithm. If $\ln y = 2$, then $y = e^2$. Therefore $\frac{2x+1}{x-3} = e^2$.

Now solve this equation. Multiply both sides by $x-3$ to clear the denominator, giving $2x+1 = e^2 x - 3e^2$. Bring all terms involving $x$ to one side: $2x - e^2 x = -3e^2 - 1$. Factor the left side as $x(2-e^2)$, then divide by $2-e^2$ to get $x = \frac{-3e^2 - 1}{2-e^2}$. Multiplying numerator and denominator by $-1$ gives the cleaner form $x = \frac{3e^2+1}{e^2-2}$.

Finally, check the domain. The original logarithms require $x-3>0$ and $2x+1>0$, so $x>3$. Since $e^2>2$, the denominator $e^2-2$ is positive, and the numerator is greater than $3(e^2-2)$, so indeed $x>3$. The solution is valid.

Key Takeaways

The question tests two central ideas: the logarithm law $\ln a - \ln b = \ln(a/b)$, and the inverse relationship between $\ln$ and $e^x$. It also tests solving a linear equation after clearing a denominator. Always remember to check that a logarithmic solution lies in the domain of the original expression.

Common Mistakes

A common mistake is writing $\ln(2x+1) - \ln(x-3) = \ln(2x+1 - (x-3))$; subtraction of logs means division of arguments, not subtraction. Another is exponentiating only one side, or writing $e^{\ln(2x+1) - \ln(x-3)} = e^2$ incorrectly. Also, when multiplying by $x-3$, students sometimes forget to multiply the whole right-hand side, giving $2x+1 = e^2 x - 3$ instead of $e^2 x - 3e^2$. Finally, sign errors in collecting $x$-terms are common.

Things to Be Careful About

The domain of the original equation is $x>3$, since both $x-3$ and $2x+1$ must be positive. The final value must satisfy this. Also, $x-3$ cannot be zero. The answer may be written in equivalent forms, such as $\frac{-3e^2-1}{2-e^2}$ or $\frac{3e^2+1}{e^2-2}$; the mark scheme accepts equivalent forms. Do not stop before checking that the denominator $e^2-2$ is not zero (it is not, since $e^2>2$).