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 08:57, 31 May 2018
Problem
Consider the graphs of and . We may say that:
Solution
Substitute into the other equation. Using a logarithm property, Subtract both sides by to get 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 | |
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All AHSME Problems and Solutions |
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