log
y(x) + log
x(y) = 43
What are the values of x and y?
logy(x) + logx(y) = 43
Let,logx(y) = p, so that:
p + 1/p = 43
Or, p^2 - 43p+1 = 0
Or, p = (43 +/- sqrt(1845))/2
= (43 +/-3*sqrt(205))/2
Or, logx(y)
= (43 +/-3*sqrt(205))/2
= 42.9767, 0.0233
So, y = x^42.9767, x^0.0233
Hence, the given equation admits of an infinite
number of solutions corresponding to infinite
choices of x.
Edited on August 21, 2007, 2:56 pm