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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.

an4s
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S.Ban
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1 Answers1

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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.

zipirovich
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  • Hey, just following up on this. So this means that my simultaneous equations will look like: ln(25/e2)=aln(5+k) eāˆ’5=aln(x+5)+k right? ā€“ S.Ban May 27 '18 at 03:19
  • @S.Ban: Close, but not exactly. In the first equation you made a typo: the right-hand side must be $a\ln(5)+k$. As for the $x$-intercept, I've just noticed (missed that when posting my original answer) that you're setting it up incorrectly. The given $x$-intercept means that the graph passes thru the point $(e-5,0)$, while you incorrectly set up something totally different. ā€“ zipirovich May 27 '18 at 13:03