Is there any algebraic solution to the equation:
$\displaystyle
\frac{e^x-e^0}{x-0}=50 ?
$
Regards
Nope. See this reference for a nice, technical examination of the problem: https://cs.uwaterloo.ca/research/tr/1993/03/W.pdf
You are asking about $\displaystyle e^x= 50x$. Rewrite it as $\displaystyle xe^{-x}= \frac{1}{50}$. Let y= -x. Then x= -y and the equation becomes $\displaystyle -ye^y= \frac{1}{50}$ or $\displaystyle ye^y= -\frac{1}{50}$. Now apply the Lambert W function (which is defined as the inverse function to $\displaystyle f(x)= xe^x$) to both sides: $\displaystyle y= W\left(-\frac{1}{50}\right)$ so $\displaystyle x= -W\left(-\frac{1}{50}\right)$.