Every root makes b to either side the same, so it is a root of
P_1^(k_1 L) ... P_n^(k_n L) = b^(c L), for L the common denominator of the
k_i: a polynomial equation, with finitely many roots, which holds at every root of the
original. The converse does not hold. Over the complex numbers log(u) + log(v) is
log(u v) only up to a multiple of 2 pi i, and a factor can be zero, so each
candidate is put back into the equation as written, under the principal logarithm, and kept
only where it holds. ln(x) + ln(x + 1) = 0 is x (x + 1) = 1, whose roots are
(-1 +- sqrt(5)) / 2; at the negative one both logarithms carry i pi and their sum
is 2 pi i, so the answer is { (sqrt(5) - 1) / 2 }. Newton's method answered it
in decimals, and ln(1 + 3x^2) - ln(4x) = 1 with one of its two roots.