Difference between revisions of "1980 AHSME Problems/Problem 18"
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== Solution == | == Solution == | ||
| − | < | + | Since <math>\log_b \sin x = a</math>, <math>b^a=\sin x</math>. |
| − | < | + | |
| − | < | + | Let <math>c = \log_b \cos x</math>. Then <math>b^c=\cos x</math>. |
| − | < | + | |
Since <math>\sin^2x+\cos^2x=1</math>, | Since <math>\sin^2x+\cos^2x=1</math>, | ||
<cmath>(b^c)^2+(b^a)^2=1</cmath> | <cmath>(b^c)^2+(b^a)^2=1</cmath> | ||
| Line 15: | Line 15: | ||
<cmath>b^{2c}=1-b^{2a}</cmath> | <cmath>b^{2c}=1-b^{2a}</cmath> | ||
<cmath>\log_b (1-b^{2a}) = 2c</cmath> | <cmath>\log_b (1-b^{2a}) = 2c</cmath> | ||
| − | <cmath>c=\boxed{\ | + | <cmath>c=\boxed{(\textbf{D}) \ \frac 12 \log_b(1-b^{2a})}</cmath> |
-aopspandy | -aopspandy | ||
Latest revision as of 13:15, 15 August 2025
Problem
If 
, 
, 
, and 
, then 
equals
Solution
Since , 
.
Let 
. Then 
.
Since 
,
![\[(b^c)^2+(b^a)^2=1\]](http://latex.artofproblemsolving.com/6/0/5/605e9e508bc763c5ea257396727557b6681839fa.png)
![\[b^{2c}+b^{2a}=1\]](http://latex.artofproblemsolving.com/f/7/b/f7b34ab01d51fdafb59c1d1b2e10f8c3238ea83f.png)
![\[b^{2c}=1-b^{2a}\]](http://latex.artofproblemsolving.com/3/3/6/3366d9834575c0c37048035fc716f6c8118c0802.png)
![\[\log_b (1-b^{2a}) = 2c\]](http://latex.artofproblemsolving.com/a/f/b/afb0b5899af5cfbb6d3de2eb1317a702ae3883f5.png)
![\[c=\boxed{(\textbf{D}) \ \frac 12 \log_b(1-b^{2a})}\]](http://latex.artofproblemsolving.com/c/2/d/c2ddd72068cebf8903fa410c904930a0990a43ed.png)
-aopspandy
See also
| 1980 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
