thanks
The line tangent to a function $\displaystyle f(x)$ at $\displaystyle x=c$ (assuming the derivative exists at this point) is found by the point slope formula:
$\displaystyle y-f(c)=f'(c)(x-c)$
Writing this in slope-intercept form, we have:
$\displaystyle y=f'(c)x+(f(c)-cf'(c))$
Now, using $\displaystyle f(x)=\frac{1}{\ln(bx)}$ and $\displaystyle c=\frac{e}{b}$ can you now compute the required $\displaystyle y$-intercept?
Your derivative is correct, and next you correctly cite the point-slope formula for a line, but after that you contradict yourself/make wrong statements, and I am not sure what you really mean. I think what I stated is about as simple as it gets.