Mock AIME II 2012 Problems/Problem 12
Revision as of 02:19, 5 April 2012 by Testingtesting (talk | contribs) (Created page with "==Problem== Let <math>\log_{a}b=5\log_{b}ac^4=3\log_{c}a^2b</math>. Assume the value of <math>\log_ab</math> has three real solutions <math>x,y,z</math>. If <math>\frac{1}{x}+\fr...")
Problem
Let 
. Assume the value of 
has three real solutions 
. If 
, where 
and 
are relatively prime positive integers, find 
.
Solution
Let 
. Then 
and 
. From this, we have the system
![\[a^{15x}=b\]](http://latex.artofproblemsolving.com/d/1/f/d1faebbba8451917acc776e916baddab70672794.png)
![\[b^{3x}=ac^4\]](http://latex.artofproblemsolving.com/7/4/b/74bf320f9d78e73fa4fd8f09160dce6b1c0c52a9.png)
![\[c^{5x}=a^2b\]](http://latex.artofproblemsolving.com/4/2/1/421759d76d8ada823e20dedef1ede488550f4399.png)
Substituting the first equation into the second, we obtain
![\[a^{45x^2}=ac^4\rightarrow a^{\frac{45x^2-1}{4}}=c\]](http://latex.artofproblemsolving.com/6/f/5/6f54e572160a2e35b85802a17a0710c393dff728.png)
Plugging this into the third equation yields 
.
Thus, 
. Note that our three real roots multiply to 
.
However, since 
, we need to multiply by 
, so our 
is
![\[\frac{8}{225}\cdot 15^3=8\cdot 15=120\]](http://latex.artofproblemsolving.com/8/c/3/8c3781cd3399442b3336b05c805edae742d1965a.png)
We need 
. Using vieta’s and making sure we count for each factor of 
we divided off, we have 
.
Our answer is 
, thus 
.
