Difference between revisions of "1961 AHSME Problems/Problem 19"
Rockmanex3 (talk | contribs) (Solution to Problem 19) |
Rockmanex3 (talk | contribs) m (→Solution) |
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| Line 14: | Line 14: | ||
Using a logarithm property, | Using a logarithm property, | ||
<cmath>2\log{x} = \log{2} + \log{x}</cmath> | <cmath>2\log{x} = \log{2} + \log{x}</cmath> | ||
| − | Subtract both sides by \log{x} to get | + | Subtract both sides by <math>\log{x}</math> to get |
<cmath>\log{x}=\log{2}</cmath> | <cmath>\log{x}=\log{2}</cmath> | ||
Thus, <math>x=2</math>. Since both <math>2\log{x}</math> and <math>\log{2x}</math> are functions, there is only one corresponding y-value, so the answer is <math>\boxed{\textbf{(B)}}</math>. | Thus, <math>x=2</math>. Since both <math>2\log{x}</math> and <math>\log{2x}</math> are functions, there is only one corresponding y-value, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
Latest revision as of 09:57, 31 May 2018
Problem
Consider the graphs of 
and 
. We may say that:
Solution
Substitute 
into the other equation.
![\[2\log{x} = \log{2x}\]](http://latex.artofproblemsolving.com/c/7/7/c7752d79d040b10dd338e894a96f60fa4525a359.png)
Using a logarithm property,
![\[2\log{x} = \log{2} + \log{x}\]](http://latex.artofproblemsolving.com/a/6/4/a641a1ad4f8ab424310a56b6f6b8b656bd5b6e28.png)
Subtract both sides by 
to get
![\[\log{x}=\log{2}\]](http://latex.artofproblemsolving.com/3/0/3/303a31d6e7e836b14094703bac06c19b55cfffc5.png)
Thus, 
. Since both 
and 
are functions, there is only one corresponding y-value, so the answer is 
.
See Also
| 1961 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
