Date | November 2014 | Marks available | 3 | Reference code | 14N.1.sl.TZ0.4 |
Level | SL only | Paper | 1 | Time zone | TZ0 |
Command term | Write | Question number | 4 | Adapted from | N/A |
Question
Write the expression 3ln2−ln4 in the form lnk, where k∈Z.
Hence or otherwise, solve 3ln2−ln4=−lnx.
Markscheme
correct application of lnab=blna (seen anywhere) (A1)
egln4=2ln2, 3ln2=ln23, 3log2=log8
correct working (A1)
eg3ln2−2ln2, ln8−ln4
ln2(accept k=2) A1 N2
[3 marks]
METHOD 1
attempt to substitute their answer into the equation (M1)
egln2=−lnx
correct application of a log rule (A1)
egln1x, ln12=lnx, ln2+lnx=ln2x(=0)
x=12 A1 N2
METHOD 2
attempt to rearrange equation, with 3ln2 written as ln23 or ln8 (M1)
eglnx=ln4−ln23, ln8+lnx=ln4, ln23=ln4−lnx
correct working applying lna±lnb (A1)
eg48, 8x=4, ln23=ln4x
x=12 A1 N2
[3 marks]
Total [6 marks]
Examiners report
Part (a) was answered correctly by a large number of candidates, though there were quite a few who applied the rules of logarithms in the wrong order.
In part (b), many candidates knew to set their answer from part (a) equal to −lnx, but then a good number incorrectly said that ln2=−lnx led to 2=−x.