Determine all possible pair(s) (p, q) of nonzero real numbers that satisfy this system of equations:

(3*p)^{log 3} = (7*q)^{log 7}, and:

7^{log p} = 3^{log q}

Note: All logarithms are considered in base ten.

(3p)^log(3)=(7q)^log(7)
log(3)*log(3p)=log(7)*log(7q)
log(3)^2+log(3)*log(p)=log(7)^2+log(7)*log(q)  (1)

7^log(p)=3^log(q)
log(p)*log(7)=log(q)*log(3)
log(p)=log(q)*log(3)/log(7)
substituting into (1) we get
log(3)^2+log(3)^2*log(q)/log(7)=log(7)^2+log(7)*log(q)
log(7)*log(3)^2+log(3)^2*log(q)=log(7)^3+log(7)^2*log(q)
log(7)*(log(3)^2-log(7)^2)=(log(7)^2-log(3)^2)*log(q)
thus
log(q)=-log(7)=log(1/7)
thus
log(p)=-log(7)*log(3)/log(7)=-log(3)=log(1/3)

thus
p=1/3
q=1/7

Posted by Daniel on 2011-05-30 10:37:07