Post History
#3: Post edited
by
Peter Taylor
·
2021-10-27T08:47:54Z (about 3 years ago)
I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 = \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.- (Of course, this doesn't exclude other solutions: see "*if rigour is not required*" above).
- I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.
- (Of course, this doesn't exclude other solutions: see "*if rigour is not required*" above).
#2: Post edited
by
Peter Taylor
·
2021-10-27T08:47:41Z (about 3 years ago)
I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 = \frac{q^{12}}{q^8} \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.- (Of course, this doesn't exclude other solutions: see "*if rigour is not required*" above).
- I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 = \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.
- (Of course, this doesn't exclude other solutions: see "*if rigour is not required*" above).
#1: Initial revision
by
Peter Taylor
·
2021-10-27T08:46:57Z (about 3 years ago)
I think your approach is the expected one, but a shortcut if rigour is not required would be to note that the absolute value of the ratio must be greater than 1, or the proportion couldn't exceed $\frac{12}8$; but then the largest term dominates, so $$q^4 = \frac{q^{12}}{q^8} \approx \frac{819}{51} \approx 16.05$$ and then trial and error shows that $q^4 = 16$ works.
(Of course, this doesn't exclude other solutions: see "*if rigour is not required*" above).