Difference between revisions of "1978 AHSME Problems/Problem 17"
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Latest revision as of 11:56, 3 June 2024
Problem 17
If 
is a positive number and 
is a function such that, for every positive number 
, ![$\left[f(x^2+1)\right]^{\sqrt{x}}=k$](http://latex.artofproblemsolving.com/1/6/b/16b093f6996ab4f6c165f7f69a7bf2cc2554237b.png)
;
then, for every positive number 
, ![$\left[f(\frac{9+y^2}{y^2})\right]^{\sqrt{\frac{12}{y}}}$](http://latex.artofproblemsolving.com/f/f/f/fff8c3d2cf5f6fb304c64e9f023193fbfb480fca.png)
is equal to
Solution
We are given that ![\[[f(x^2 + 1)]^{\sqrt(x)} = k\]](http://latex.artofproblemsolving.com/e/9/4/e94b869a0432519319f754ebb1aa2a49729f295f.png)
We can rewrite 
as 
Thus, our function is now
![\[[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{12}{y}}} = k\]](http://latex.artofproblemsolving.com/2/4/a/24a6b6521e197bec65a6e9ed056bfa3649fd0e86.png)
![\[\Rrightarrow[f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y} \cdot 4}} = k\]](http://latex.artofproblemsolving.com/7/0/b/70b8fd9d7af0af5e85b2e7b9f0524a21554d089b.png)
![\[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{\sqrt{4}} = (k)^{\sqrt{4}}\]](http://latex.artofproblemsolving.com/2/d/7/2d71e5b8e88878b025e457cc7b90cf94a36ee380.png)
![\[\Rrightarrow([f(\frac{9}{y^2} + 1)]^{\sqrt{\frac{3}{y}}})^{2} = (k)^{2} = k^2\]](http://latex.artofproblemsolving.com/0/2/e/02e460018013bb355ec9ae042c0c9e10e496d44d.png)
Thus, the answer is 
.
~JustinLee2017
~clarification by Leon
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
