# Given constants $c_i$ and $K$, find $b$ numerically such that $b^{c_1} + b^{c_2} + ... + b^{c_n} = K$

Given constants $$c_i > 0$$ and $$K > 0$$, find $$b > 0$$ numerically such that $$b^{c_1} + b^{c_2} + \dots + b^{c_n} = K$$. I'd like to solve this with a non-iterative method if possible.

My attempt at isolating b (applying the identity $$\log(x + y) = \log x + \log(1 + y/x)$$):

$$K = b^{c_1} + b^{c_2} + \dots$$ $$\log(K) = \log(b^{c_1} + b^{c_2} + \dots)$$ $$\log(K) = c_1 \log(b) + \log(1 + b^{c_2-c_1} + b^{c_3-c_1} + \dots)$$ $$\log(K) = c_1 \log(b) + \log(1 + b^{c_2-c_1}) + \log(1 + \frac{b^{c_3-c_1}}{(1 + b^{c_2-c_1})} + \dots)$$ $$\log(K) = c_1 \log(b) + \log(1 + b^{c_2-c_1}) + \log(1 + \frac{b^{c_3-c_1}}{(1 + b^{c_2-c_1})}) + \log(1 + \dots)$$

It doesn't appear to lead anywhere clean where I can isolate $$b$$ and use log1p. Any ideas?