Standard +0.3 This is a straightforward simultaneous equations problem requiring knowledge of logarithm laws (difference of logs = log of quotient) and substitution. The steps are routine: convert the log equation to a/b = 4³ = 64, substitute into ab = 25 to get a² = 1600, then solve. While it combines logs and algebra, it follows a standard template with no novel insight required, making it slightly easier than average.
6. Given that \(a\) and \(b\) are positive constants, solve the simultaneous equations
$$\begin{gathered}
a b = 25 \\
\log _ { 4 } a - \log _ { 4 } b = 3
\end{gathered}$$
Show each step of your working, giving exact values for \(a\) and \(b\).
\(\log_4 \frac{a}{b} = 3\) or \(\log_4 a + \log_4 b = \log_4 25\) or \(\log_4\frac{a}{\frac{25}{b}}=3\) or \(\log_4\frac{\frac{b}{a}}{\frac{25}{b}}=3\)
M1
M1: Uses addition or subtraction law correctly for logs. N.B. \(\log_4 a + \log_4 b = 25\) is M0
\(\log_4 64 = 3\) or \(4^3 = 64\) (may be implied by use of 64) or \(\log a = \frac{1}{2}(\log 25+3)\) becoming \(a = 4^{\frac{1}{2}(\log 25+3)}\)
B1
B1: See 64 used independently of M mark
Correct algebraic elimination of a variable to obtain expression in \(a\) or \(b\) without logs
dM1
dM1: Dependent on first M. Eliminates \(a\) or \(b\) with appropriate algebra
\(a = 40\) or \(b = \frac{5}{8}\)
A1
A1: Either \(a\) or \(b\) correct
Substitutes to give second variable or solves again from start
dM1
dM1: Dependent on first M
\(a = 40\) and \(b = \frac{5}{8}\) and no other answers
A1
A1: Both correct – allow \(b=0.625\). If \(a=-40\) and \(b=-\frac{5}{8}\) also given, lose last A mark. NB rounding to 40 and 0.625 after decimal work: do not give final A mark
[6]
# Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_4 \frac{a}{b} = 3$ or $\log_4 a + \log_4 b = \log_4 25$ or $\log_4\frac{a}{\frac{25}{b}}=3$ or $\log_4\frac{\frac{b}{a}}{\frac{25}{b}}=3$ | M1 | M1: Uses addition or subtraction law correctly for logs. N.B. $\log_4 a + \log_4 b = 25$ is M0 |
| $\log_4 64 = 3$ or $4^3 = 64$ (may be implied by use of 64) or $\log a = \frac{1}{2}(\log 25+3)$ becoming $a = 4^{\frac{1}{2}(\log 25+3)}$ | B1 | B1: See 64 used independently of M mark |
| Correct algebraic elimination of a variable to obtain expression in $a$ or $b$ without logs | dM1 | dM1: Dependent on first M. Eliminates $a$ or $b$ with appropriate algebra |
| $a = 40$ or $b = \frac{5}{8}$ | A1 | A1: Either $a$ or $b$ correct |
| Substitutes to give second variable or solves again from start | dM1 | dM1: Dependent on first M |
| $a = 40$ **and** $b = \frac{5}{8}$ and no other answers | A1 | A1: Both correct – allow $b=0.625$. If $a=-40$ and $b=-\frac{5}{8}$ also given, **lose last A mark**. NB rounding to 40 and 0.625 after decimal work: do not give final A mark |
| | [6] | |
6. Given that $a$ and $b$ are positive constants, solve the simultaneous equations
$$\begin{gathered}
a b = 25 \\
\log _ { 4 } a - \log _ { 4 } b = 3
\end{gathered}$$
Show each step of your working, giving exact values for $a$ and $b$.\\
\hfill \mbox{\textit{Edexcel C12 2014 Q6 [6]}}