1978 AHSME Problems/Problem 17
Revision as of 03:12, 13 February 2021 by Justinlee2017 (talk | contribs) (Created page with "We are given that <cmath>[f(x^2 + 1)]^{\sqrt(x)} = k</cmath> We can rewrite <math>\frac{9+y^2}{y^2}</math> as <math>\frac{9}{y^2} + 1</math> Thus, our function is now <cmath>[...")
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)
![\[\boxed{D}\]](http://latex.artofproblemsolving.com/d/a/d/dadc8e71363a80c53a137421994b9e02de6eeabd.png)
~JustinLee2017
