Difference between revisions of "1956 AHSME Problems/Problem 20"
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Taking reciprocals of the first condition yields <math>5^x=\frac{1}{2}</math>. Then <math>5^x\cdot2^x=\frac{1}{2}\cdot2^x</math>, so <math>10^x=2^{x-1}</math>. Taking base <math>10</math> log of both sides results in <math>x=(x-1)\log 2</math>, so | Taking reciprocals of the first condition yields <math>5^x=\frac{1}{2}</math>. Then <math>5^x\cdot2^x=\frac{1}{2}\cdot2^x</math>, so <math>10^x=2^{x-1}</math>. Taking base <math>10</math> log of both sides results in <math>x=(x-1)\log 2</math>, so | ||
<cmath>x=\frac{\log2}{\log2-1}\approx{0.3}{-0.7}=-\frac{3}{7}\approx-0.429</cmath> | <cmath>x=\frac{\log2}{\log2-1}\approx{0.3}{-0.7}=-\frac{3}{7}\approx-0.429</cmath> | ||
| − | Thus <math>\boxed{\textbf{( | + | Thus <math>\boxed{\textbf{(C)}}</math> is correct. |
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406] | ||
Latest revision as of 21:23, 19 March 2025
If 
and 
, then the value of 
to the nearest tenth is:
Solution
Taking reciprocals of the first condition yields 
. Then 
, so 
. Taking base 
log of both sides results in 
, so
![\[x=\frac{\log2}{\log2-1}\approx{0.3}{-0.7}=-\frac{3}{7}\approx-0.429\]](http://latex.artofproblemsolving.com/d/6/8/d6886bcdef3102b8d8fb5bd996e1a0d74605f400.png)
Thus 
is correct.
See also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
