There was a question that I came across that I was unable to answer..
Find the equation to a logarithmic equation with an asymptote at $x=-5$ and containing the $X$-intercept $X=e-5$ and $Y$-intercept $Y=\log_e\dfrac{25}{e^2}$.
I understand that the basic formula of log graphs is $$y=a\log_b(x-h)+k$$ and that given the horizontal asymptote at $x = -5$ $$y=a\log_b(x+5)+k$$
I also understand that we should be able to have two simultaneous equations: $$e-5=a\log_b(x+5)+k$$ and $$\log_e\dfrac{25}{e^2}=a\log_b5+k$$
But I am unsure of where to go from here in order to get the full equation.
Recall all logarithms are proportional to each other via the change of base formula:
$$\log_b(a)=\frac{\log_c(a)}{\log_c(b)}; \quad \text{in particular } \log_b(a)=\frac{\ln(a)}{\ln(b)}.$$
Therefore, you can set up the original function in the form $$y=a\ln(x-h)+k,$$ because the coefficient $a$ in front of the natural logarithm is equivalent to switching to other bases. Now your approach is going to work.