Difference between revisions of "1981 AHSME Problems/Problem 15"
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Use rules of logarithms to solve this equation. | Use rules of logarithms to solve this equation. | ||
| − | < | + | <cmath>(2x)^{\log_{b} 2} = (3x)^{\log_{b} 3}</cmath> |
| − | < | + | <cmath>\log_{b} 2 \cdot \log_{b} 2x = \log_{b} 3 \cdot \log_{b} 3x</cmath> |
| − | < | + | <cmath>\log_{b} 2 \cdot (\log_{b} 2 + \log_{b} x) = \log_{b} 3 \cdot (\log_{b} 3 + \log_{b} x)</cmath> |
| − | < | + | <cmath>(\log_{b} 2)^2 + \log_{b} 2 \cdot \log_{b} x = (\log_{b} 3)^2 + \log_{b} 3 \cdot \log_{b} x</cmath> |
| − | < | + | <cmath>(\log_{b} 2)^2 - (\log_{b} 3)^2 = \log_{b} 3 \cdot \log_{b} x - \log_{b} 2 \cdot \log_{b} x </cmath> |
| − | < | + | <cmath>(\log_{b} 2 + \log_{b} 3)(\log_{b} 2 - \log_{b} 3) = \log_{b} x \cdot (\log_{b} 3 - \log_{b} 2)</cmath> |
| + | |||
| + | <cmath>\log_{b} 2 + \log_{b} 3 = -\log_{b} x</cmath> | ||
| + | |||
| + | <cmath>\log_{b} 6 = -\log_{b} x</cmath> | ||
| + | |||
| + | <cmath>6 = \frac{1}{x}</cmath> | ||
| + | |||
| + | <cmath>x = \boxed {(\textbf{B})\ \frac{1}{6}}</cmath> | ||
| + | |||
| + | -j314andrews | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AHSME box|year=1981|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 16:57, 20 August 2025
Problem
If 
, 
, and 
, then 
is
Solution
Use rules of logarithms to solve this equation.
![\[(2x)^{\log_{b} 2} = (3x)^{\log_{b} 3}\]](http://latex.artofproblemsolving.com/8/4/5/845e162fbf47141d2442a3975785f4382857acfc.png)
![\[\log_{b} 2 \cdot \log_{b} 2x = \log_{b} 3 \cdot \log_{b} 3x\]](http://latex.artofproblemsolving.com/e/2/e/e2eb649b11eaca610b77af875f6d40cf035ea462.png)
![\[\log_{b} 2 \cdot (\log_{b} 2 + \log_{b} x) = \log_{b} 3 \cdot (\log_{b} 3 + \log_{b} x)\]](http://latex.artofproblemsolving.com/3/f/4/3f4e59028a7c0990667dac81817bdaed3140ce79.png)
![\[(\log_{b} 2)^2 + \log_{b} 2 \cdot \log_{b} x = (\log_{b} 3)^2 + \log_{b} 3 \cdot \log_{b} x\]](http://latex.artofproblemsolving.com/4/a/6/4a63882e9b417f4f85eaf3ceb58ca21f1b1957dc.png)
![\[(\log_{b} 2)^2 - (\log_{b} 3)^2 = \log_{b} 3 \cdot \log_{b} x - \log_{b} 2 \cdot \log_{b} x\]](http://latex.artofproblemsolving.com/e/0/5/e05d1a6b1c8309c7fe6bab2947a5ecc1dd2e61b2.png)
![\[(\log_{b} 2 + \log_{b} 3)(\log_{b} 2 - \log_{b} 3) = \log_{b} x \cdot (\log_{b} 3 - \log_{b} 2)\]](http://latex.artofproblemsolving.com/f/6/7/f678d0fadfaeaffb6aca8c2ee89eee9cc65fd702.png)
![\[\log_{b} 2 + \log_{b} 3 = -\log_{b} x\]](http://latex.artofproblemsolving.com/3/1/0/31032e3cdf0e9c8148a509d74e7f8a5d5830a08e.png)
![\[\log_{b} 6 = -\log_{b} x\]](http://latex.artofproblemsolving.com/8/1/0/810f5c0415ce5db08247412fc55b20879a50c2b2.png)
![\[6 = \frac{1}{x}\]](http://latex.artofproblemsolving.com/0/0/4/00438f3c9ebb157e62315fc6c51c9595953f56b2.png)
![\[x = \boxed {(\textbf{B})\ \frac{1}{6}}\]](http://latex.artofproblemsolving.com/9/3/8/93877b7c27d9cecb7370ea3f819c9b944963531c.png)
-j314andrews
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
