Standard +0.3 This is a straightforward simultaneous equations problem combining logarithm laws with algebraic manipulation. Students need to apply log₅(a) + log₅(b) = log₅(ab) = 3 to get ab = 125, then solve the linear-quadratic system using substitution. While it requires multiple steps and surd simplification, the techniques are standard P2 material with no novel insight required, making it slightly easier than average.
In this question you must show all stages of your working.
Give your answers in fully simplified surd form.
Given that \(a\) and \(b\) are positive constants, solve the simultaneous equations
\begin{align}
a - b &= 8
\log_5 a + \log_5 b &= 3
\end{align}
[6]
M1: Correct addition law applied (may be implied) or undoes a log equation correctly, or
replaces the 3 by l o g 4 3 oe
4
The addition law may be seen after substitution from the first equation e.g.
a − b = 8 a = b + 8 l o g a + l o g b = l o g ( b + 8 ) + l o g b = l o g b ( b + 8 )
4 4 4 4 4
Condone l o g ( b + 8 ) + l o g b = l o g b 2 + 8 b
4 4 4
dM1: Removes logs correctly and proceeds from an equation of the form ab= K and the
given equation a−b=8 to a quadratic equation in either a or b.
A1: Correct quadratic equation. Brackets must be expanded but terms not necessarily all on one
side. The “ =0” may be implied by their attempt to solve.
M1: Solves a 3TQ in a or b to obtain a positive solution. There is no need to see the
negative solution in their working.
M1: Uses their a to find b or vice versa. This depends on having solved a 3TQ earlier and obtained a
root of the form p + q r , p , q , r 0 to obtain another root of a similar form i.e. not a decimal.
The value of a or b does not necessarily have to be positive for this mark.
Note that having found a or b they may repeat the process above to find the other value which
is acceptable.
A1: Both of a = 4 + 4 5 and b = − 4 + 4 5 simplified, with no other solutions.
Apply isw once correct answers are seen, e.g. if they subsequently go into decimals.
Answer
Marks
Guidance
Question
Scheme
Marks
Question 4:
4 | l o g a + l o g b = l o g a b or l o g " a b " = 3 " a b " = 4 3
4 4 4 4 | M1
a b = K , a − b = 8 a ( a − 8 ) = K o r ( b + 8 ) b = K | dM1
a2 −8a−64=0 or b2 +8b−64=0 | A1
− ( − 8 ) + ( − 8 ) 2 − 4 1 − 6 4 − 8 + 8 2 − 4 1 − 6 4
a = = ... or b = = . . .
2 1 2 1 | M1
a = ' 4 + 4 5 ' b=−4+4 5 | M1
a=4+4 5 and b=−4+4 5 and no other solutions. | A1
(6)
(6 marks)
Notes:
M1: Correct addition law applied (may be implied) or undoes a log equation correctly, or
replaces the 3 by l o g 4 3 oe
4
The addition law may be seen after substitution from the first equation e.g.
a − b = 8 a = b + 8 l o g a + l o g b = l o g ( b + 8 ) + l o g b = l o g b ( b + 8 )
4 4 4 4 4
Condone l o g ( b + 8 ) + l o g b = l o g b 2 + 8 b
4 4 4
dM1: Removes logs correctly and proceeds from an equation of the form ab= K and the
given equation a−b=8 to a quadratic equation in either a or b.
A1: Correct quadratic equation. Brackets must be expanded but terms not necessarily all on one
side. The “ =0” may be implied by their attempt to solve.
M1: Solves a 3TQ in a or b to obtain a positive solution. There is no need to see the
negative solution in their working.
M1: Uses their a to find b or vice versa. This depends on having solved a 3TQ earlier and obtained a
root of the form p + q r , p , q , r 0 to obtain another root of a similar form i.e. not a decimal.
The value of a or b does not necessarily have to be positive for this mark.
Note that having found a or b they may repeat the process above to find the other value which
is acceptable.
A1: Both of a = 4 + 4 5 and b = − 4 + 4 5 simplified, with no other solutions.
Apply isw once correct answers are seen, e.g. if they subsequently go into decimals.
Question | Scheme | Marks
In this question you must show all stages of your working.
Give your answers in fully simplified surd form.
Given that $a$ and $b$ are positive constants, solve the simultaneous equations
\begin{align}
a - b &= 8\\
\log_5 a + \log_5 b &= 3
\end{align}
[6]
\hfill \mbox{\textit{Edexcel P2 2022 Q4 [6]}}