Moderate -0.8 This is a straightforward exponential equation requiring standard logarithm techniques: take ln of both sides, apply log laws (ln(a^b) = b ln a and ln(3^{x+4}) = (x+4)ln 3), rearrange to collect x terms, and factor. The 'show that' format removes problem-solving demand, and the algebraic manipulation is routine for A-level. Slightly easier than average due to its mechanical nature and clear path to solution.
6 Show that the solution of the equation
$$5 ^ { x } = 3 ^ { x + 4 }$$
can be written as
$$x = \frac { \ln 81 } { \ln 5 - \ln 3 }$$
Fully justify your answer.
Takes logs of both sides and uses a log rule correctly
\(x\ln 5=(x+4)\ln 3\), so \(x\ln 5 - x\ln 3 = 4\ln 3\), so \(x(\ln 5-\ln 3)=\ln 81\)
M1 (1.1a)
Applies all necessary log rules so \(x\) is no longer an exponent and expresses \(4\ln 3\) in terms of \(x\); condone sign error
\(\ln 81\) from \(4\ln 3\) or from \(3^x \times 3^4\)
B1 (1.1b)
\(x=\frac{\ln 81}{\ln 5 - \ln 3}\)
R1 (2.1)
Completes reasoned argument; must see \(x(\ln 5-\ln 3)\) on penultimate line; if natural logs not used throughout, base must be converted at end
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\ln 5^x = \ln 3^{x+4}$ | M1 (1.1a) | Takes logs of both sides and uses a log rule correctly |
| $x\ln 5=(x+4)\ln 3$, so $x\ln 5 - x\ln 3 = 4\ln 3$, so $x(\ln 5-\ln 3)=\ln 81$ | M1 (1.1a) | Applies all necessary log rules so $x$ is no longer an exponent and expresses $4\ln 3$ in terms of $x$; condone sign error |
| $\ln 81$ from $4\ln 3$ or from $3^x \times 3^4$ | B1 (1.1b) | |
| $x=\frac{\ln 81}{\ln 5 - \ln 3}$ | R1 (2.1) | Completes reasoned argument; must see $x(\ln 5-\ln 3)$ on penultimate line; if natural logs not used throughout, base must be converted at end |
6 Show that the solution of the equation
$$5 ^ { x } = 3 ^ { x + 4 }$$
can be written as
$$x = \frac { \ln 81 } { \ln 5 - \ln 3 }$$
Fully justify your answer.\\
\hfill \mbox{\textit{AQA Paper 2 2021 Q6 [4]}}