Date | November 2021 | Marks available | 5 | Reference code | 21N.1.AHL.TZ0.3 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Solve | Question number | 3 | Adapted from | N/A |
Question
Solve the equation log3 √x=12 log2 3+log3(4x3), where x>0.
Markscheme
attempt to use change the base (M1)
log3 √x=log3 22+log3(4x3)
attempt to use the power rule (M1)
log3 √x=log3 √2+log3(4x3)
attempt to use product or quotient rule for logs, ln a+ln b=ln ab (M1)
log3 √x=log3 (4√2x3)
Note: The M marks are for attempting to use the relevant log rule and may be applied in any order and at any time during the attempt seen.
√x=4√2x3
x=32x6
x5=132 (A1)
x=12 A1
[5 marks]
Examiners report
[N/A]