Difference between revisions of "1951 AHSME Problems/Problem 45"
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Latest revision as of 11:27, 5 July 2013
Problem 45
If you are given 
and 
, then the only logarithm that cannot be found without the use of tables is:
Solution
While 
, we cannot easily deal with the logarithm of a sum. Furthermore, 
is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore 
.
As for the rest of the cases: ![\[\log\frac{5}{4} = \log\frac{10}{8} = \log 10 - \log 8 = 1 - \log 8\]](http://latex.artofproblemsolving.com/4/4/a/44a818e38e926d62fd48dc0a60236b2966c5ea57.png)
can be found; ![\[\log 15 = \log 3 + \log 5 = \frac{1}{2}\log 3^2 + \log\frac{10}{2} = \frac{1}{2} \log 9 + \log 10 - \log 2 = \frac{1}{2} \log 9 + 1 - \frac{1}{3} \log 8\]](http://latex.artofproblemsolving.com/8/d/f/8df7140df5de2bcea8fa51f0a10e54bb1fca686f.png)
can be found; ![\[\log 600 = \log 100 + \log 6 = 2 + \log 2 + \log 3 = 2 + \frac{1}{3} \log 8 + \frac{1}{2} \log 9\]](http://latex.artofproblemsolving.com/0/8/1/081aee24fd2ac4aaff4d751089188837c6a70a61.png)
can be found; and ![\[\log .4 = \log\frac{4}{10} = \log 4 - \log 10 = 2 \log 2 - 1 = \frac{2}{3} \log 8 - 1\]](http://latex.artofproblemsolving.com/3/9/7/397bb0db12625f927b748903f2f78c7f3871215f.png)
can be found.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 44 |
Followed by Problem 46 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
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