Solve $$ y=\log_{b}\left(1+b^{x}\right) $$ or the equivalent $$ b^{y}-b^{x}=1 $$ for $ b $
if it helps, $b>0$ and $y>0$
This can model the value of a call option (ignoring time) as a function of the stock price with $$ x = StockPrice - StrikePrice $$ $$ y = OptionValue $$
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The equation $$b^y-b^x=1\tag 1$$ cannot be solved in the most general case, even using special functions such as Lambert.
If $y >x$, let $t=b^x$ to make the equation $$t^{\frac y x}-t=1$$ If $\frac y x$ is a small integer $(\leq 4)$, it will be a polynomial in $t$ which could be solved.Otherwise, you would need some numerical method to solve $(1)$ and Newton is probably the simplest to use.
If $y <x$, let $t=b^y$ to make the equation $$-t^{\frac x y}+t=1$$
Considering the first case $y>x$, let $k=\frac y x$ and consider that you look for the zero of function $$f(t)=t^k-t-1$$ and Newton iterates will be given by $$t_{n+1}=t_n-\frac{t_n^k-t_n-1}{k t_n^{k-1}-1}$$ Since $f(1)=-1$, use $t_0=1$.
Trying for $k=\pi$, the iterates would be $$\left( \begin{array}{cc} n & t_n \\ 0 & 1.00000 \\ 1 & 1.46694 \\ 2 & 1.32587 \\ 3 & 1.30483 \\ 4 & 1.30438 \end{array} \right)$$
Now, go back to $b=t^{\frac{1}{x}}$.