# Thread: Domain range and inverse of natural log function

1. ## Domain range and inverse of natural log function

Let f be the function given by f(x)=ln(x/x-1)
A) what is the domain and range of f
B) what is the inverse of f
C) what is the range of the inverse,justify

So when I started this problem I got the domain to be (-infinity,0),(1,infinity) because the restriction in the domain of 1 and 0. For the range I think it is a limit issue but I don't know how to get that and for the inverse I can't figure out how to get rid of the y-1 in the denominator after I raise both sides under base e to get rid of the ln. Thanks!

2. ## Re: Domain range and inverse of natural log function

for the range, consider the first domain interval $\displaystyle (-\infty,0)$ ... as $\displaystyle x \to -\infty$, $\displaystyle \frac{x}{x-1} \to 1$ and as $\displaystyle x \to 0^-$ , $\displaystyle \frac{x}{x-1} \to 0$ ... the range of the log function over this interval is $\displaystyle (-\infty, 0)$.

now consider the other interval of the domain $\displaystyle (1, \infty)$ ... as $\displaystyle x \to 1^+$, $\displaystyle \frac{x}{x-1} \to \infty$ and as $\displaystyle x \to \infty$ , $\displaystyle \frac{x}{x-1} \to 1$ ... the range of the log function over this interval is $\displaystyle (0, \infty)$.

I'd say the range is all reals except y = 0

(b) inverse ...

$\displaystyle x = \ln\left(\frac{y}{y-1}\right)$

$\displaystyle \frac{y}{y-1} = e^x$

$\displaystyle y = ye^x - e^x$

$\displaystyle e^x = ye^x - y$

$\displaystyle e^x = y(e^x-1)$

$\displaystyle y = \frac{e^x}{e^x - 1}$

(c) ... wouldn't the range of the inverse be the domain of the original function?

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### natural log ln inverse domain range of functions

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