Mathematics 9709/32 — May/June 2022

Walkthrough

First notice that the right-hand side has both $2x$ and $\ln 3$. The $2x$ is not a logarithm, so before using the one-to-one property of logarithms we must express it in logarithmic form. Since exponentials and logarithms are inverse functions, $2x = \ln(e^{2x})$.

Rewrite the equation as:

$$\ln(e^{2x}+3) - \ln 3 = \ln(e^{2x})$$

Now the left-hand side is a difference of two logarithms. Use the quotient law $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$ to combine them:

$$\ln\left(\frac{e^{2x}+3}{3}\right) = \ln(e^{2x})$$

Because the logarithm function is one-to-one, if $\ln A = \ln B$ then $A = B$. So equate the arguments:

$$\frac{e^{2x}+3}{3} = e^{2x}$$

Multiply both sides by 3:

$$e^{2x}+3 = 3e^{2x}$$

Rearrange to collect the exponential terms on one side:

$$3 = 2e^{2x}$$ $$e^{2x} = \frac{3}{2}$$

Take natural logarithms of both sides:

$$2x = \ln\left(\frac{3}{2}\right)$$

Divide by 2:

$$x = \frac{1}{2}\ln\left(\frac{3}{2}\right)$$

Evaluating with a calculator gives $x = 0.202732\ldots$, so correct to 3 decimal places, $x = 0.203$.

Key Takeaways

The key idea is that any term of the form $kx$ can be written as $\ln(e^{kx})$, allowing an equation containing both logs and ordinary terms to be converted entirely into logarithmic form. Once both sides are single logarithms, the one-to-one property lets us equate their arguments. The final step uses the inverse relationship between exponentials and logarithms to solve for the variable.

Common Mistakes

• Trying to exponentiate the original equation immediately, before both sides are written as single logarithms.
• Incorrectly writing $\ln(e^{2x}+3)$ as $\ln(e^{2x}) + \ln 3$. The logarithm of a sum is not the sum of logarithms.
• Making a sign error when moving $\ln 3$ to the other side.
• Forgetting the factor of $\frac{1}{2}$, which would give $x \approx 0.405$ instead of $0.203$.
• Giving the answer without showing working. The mark scheme states that an answer only with no working is awarded 0 out of 4 marks.

Things to Be Careful About

• The expression $e^{2x}+3$ is always positive, so the logarithm is defined for all real $x$; there is no domain restriction to check here.
• The solution $e^{2x} = \frac{3}{2}$ is positive, so taking logarithms is valid.
• The final answer must be given correct to 3 decimal places, so write $0.203$, not just the exact logarithmic form.