The line y = mx + 4 passes through the points (3, log2 p) and (log2 p, 4).
What are the possible values of p?
- Ap = 1 and p = 4
- Bp = 1 and p = 16
- Cp = 14 and p = 4
- Dp = 14 and p = 64
- Ep = 164 and p = 4
- Fp = 164 and p = 16
Show the answer and worked solution
answer · B
- Ap = 1 and p = 4
- Bp = 1 and p = 16
- Cp = 14 and p = 4
- Dp = 14 and p = 64
- Ep = 164 and p = 4
- Fp = 164 and p = 16
Put t = log2 p. The second point gives 4 = mt + 4, so mt = 0: either t = 0 or m = 0. If t = 0 then p = 1, and the first point (3, 0) forces 0 = 3m + 4, so m = −43, which is consistent. If instead m = 0 the line is y = 4, so the first point gives t = 4 and p = 16. The possible values are p = 1 and p = 16.