Moderate -0.3 This is a straightforward exponential equation requiring logarithms to solve. Students need to take logs of both sides and rearrange to isolate x—a standard technique practiced extensively at this level. The presence of different bases (8, 4, 5) adds minor complexity but no conceptual difficulty beyond routine manipulation. Worth 4 marks, indicating multiple steps but no novel insight required.
Use law of the logarithm of product or quotient on each side
*B1
ln83 + ln8−6x and ln4 + ln5−2x.
83
Allow for ln and ln86x – ln52x.
4
(3 − 6x)ln8 and ln4 + ln5−2x gains next DB1 as well.
Use law of logarithm of a power involving x on ONE side,
e.g. ln83 + (−)6xln8 or (3 − 6x)ln8 or 918xln2
Answer
Marks
Guidance
or ln4 − 2xln5
DB1
SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law
seen anywhere.
Obtain a correct linear equation in x, e.g.
Answer
Marks
Guidance
36xln89 18xln2ln42xln5
B1
If in decimals, allow small errors in 2nd and 3rd dp.
Obtain answer x = 0.524
B1
3dp required.
No working scores 0/4 marks.
After *B1 DB1 to correct answer with no more log working seen,
then SC B1 for x = 0.524. Maximum 3/4 possible.
Alternative Method for Question 1
Use laws of indices to get to a = b±2x or c± x in a correct form so now
Answer
Marks
Guidance
only ONE log power law required
(B2)
(83/4) and (5/83)−2x or (52/86)−x opposite sides or
(4/83) and (83/5)–2x or (86/52)–x opposite sides
83 83
Obtain a correct linear equation in x, e.g.ln 2xln
Answer
Marks
Guidance
4 5
(B1)
−2xln(5/83) or 2xln(83/5) or xln(86/52) or – xln(52/86).
SC: If B0 then allow B1 (1/4) for a correct term seen anywhere.
If in decimals, allow small errors in 2nd and 3rd dp.
Answer
Marks
Guidance
Obtain answer x = 0.524
(B1)
3dp required.
No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then
B2 and SC B1 for x = 0.524. Maximum 3/4 possible.
Answer
Marks
Guidance
Question
Answer
Marks
1
Alternative Method 2 for Question 1
Use laws of indices to get to any correct form with indices
Answer
Marks
Guidance
combined so now TWO log power laws are required
(*B1)
Allow 27 – 18x and 5–2x on opposite sides
or 29 – 18x and 22 – 4.64x on opposite sides.
Use law of logarithm of a power involving x on ONE side,
e.g. (7 – 18x)ln2 = ln5–2x or ln27 – 18x = −2xln5 or …
Answer
Marks
Guidance
Allow 7 – 18xln2 or 9 18xln2
(DB1)
e.g. (7 – 18x)ln2 or 9 18xln2 or –2xln5 or (2 – 4.64x)ln2
SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen
anywhere.
E.g. any term in *B1 shown above.
Obtain a correct linear equation in x,
Answer
Marks
Guidance
e.g. (7 – 18x)ln2 = − 2xln5 or (9 – 18x)ln2 = (2 – 4.64x)ln2
(B1)
If in decimals, allow small errors in 2nd and 3rd dp.
Obtain answer x = 0.524
(B1)
3dp required.
No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then
*B1 and SC B1 for x = 0.524. Maximum 2/4 possible.
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 1:
1 | Use law of the logarithm of product or quotient on each side | *B1 | Allow logs to any base, as well as decimals, throughout.
ln83 + ln8−6x and ln4 + ln5−2x.
83
Allow for ln and ln86x – ln52x.
4
(3 − 6x)ln8 and ln4 + ln5−2x gains next DB1 as well.
Use law of logarithm of a power involving x on ONE side,
e.g. ln83 + (−)6xln8 or (3 − 6x)ln8 or 918xln2
or ln4 − 2xln5 | DB1 | SC If *B0 DB0, then allow B1 (1/4) for a correct logarithm law
seen anywhere.
Obtain a correct linear equation in x, e.g.
36xln89 18xln2ln42xln5 | B1 | If in decimals, allow small errors in 2nd and 3rd dp.
Obtain answer x = 0.524 | B1 | 3dp required.
No working scores 0/4 marks.
After *B1 DB1 to correct answer with no more log working seen,
then SC B1 for x = 0.524. Maximum 3/4 possible.
Alternative Method for Question 1
Use laws of indices to get to a = b±2x or c± x in a correct form so now
only ONE log power law required | (B2) | (83/4) and (5/83)−2x or (52/86)−x opposite sides or
(4/83) and (83/5)–2x or (86/52)–x opposite sides
83 83
Obtain a correct linear equation in x, e.g.ln 2xln
4 5 | (B1) | −2xln(5/83) or 2xln(83/5) or xln(86/52) or – xln(52/86).
SC: If B0 then allow B1 (1/4) for a correct term seen anywhere.
If in decimals, allow small errors in 2nd and 3rd dp.
Obtain answer x = 0.524 | (B1) | 3dp required.
No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then
B2 and SC B1 for x = 0.524. Maximum 3/4 possible.
Question | Answer | Marks | Guidance
1 | Alternative Method 2 for Question 1
Use laws of indices to get to any correct form with indices
combined so now TWO log power laws are required | (*B1) | Allow 27 – 18x and 5–2x on opposite sides
or 29 – 18x and 22 – 4.64x on opposite sides.
Use law of logarithm of a power involving x on ONE side,
e.g. (7 – 18x)ln2 = ln5–2x or ln27 – 18x = −2xln5 or …
Allow 7 – 18xln2 or 9 18xln2 | (DB1) | e.g. (7 – 18x)ln2 or 9 18xln2 or –2xln5 or (2 – 4.64x)ln2
SC: If *B0 DB0 then allow B1 (1/4) for a correct term seen
anywhere.
E.g. any term in *B1 shown above.
Obtain a correct linear equation in x,
e.g. (7 – 18x)ln2 = − 2xln5 or (9 – 18x)ln2 = (2 – 4.64x)ln2 | (B1) | If in decimals, allow small errors in 2nd and 3rd dp.
Obtain answer x = 0.524 | (B1) | 3dp required.
No working scores 0/4 marks.
From the first line to correct answer with no log working seen, then
*B1 and SC B1 for x = 0.524. Maximum 2/4 possible.
4
Question | Answer | Marks | Guidance