How do you find the inverse of 5/e^x+1?

Question

How do you find the inverse of 5/e^x+1?

1 Answer

Impact of this question

2299 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-17T22:47:03

$y = - ln (\frac{x - 1}{5})$ $y=-ln((x-1)/5)$

Explanation:

Remember that $f (x)$ $f(x)$ is the inverse of $g (x)$ $g(x)$ if both $f (g (x))$ $f(g(x))$ and $g (f (x))$ $g(f(x))$ both equal $x$ $x$ .

From this, we know that when $\frac{5}{e^{x}} + 1 = f (x)$ $5/e^x+1=f(x)$ , then $\frac{5}{e^{g (x)}} + 1 = x$ $5/e^g(x)+1=x$ where $g (x)$ $g(x)$ is the inverse function of $f (x)$ $f(x)$ .
For simplicity, we will use $y$ $y$ in place of $g (x)$ $g(x)$ .

We now isolate $y$ $y$ .

$\frac{5}{e^{y}} + 1 = x$ $5/e^y+1=x$
$\frac{5}{e^{y}} = x - 1$ $5/e^y=x-1$
$5 \times e^{- y} = x - 1$ $5xxe^-y=x-1$ We use the fact that $\frac{z}{x^{y}} = z \times x^{- y}$ $z/x^y=zxx x^-y$
$e^{- y} = \frac{x - 1}{5}$ $e^-y=(x-1)/5$ We use the natural logarithm on both sides.
${log}_{e} (e^{- y}) = {log}_{e} (\frac{x - 1}{5})$ $Log _e(e^-y)=Log_e((x-1)/5)$ which is really
$ln (e^{- y}) = ln (\frac{x - 1}{5})$ $ln(e^-y)=ln((x-1)/5)$ Remember that ${log}_{x} x^{y} = y$ $log_x x^y=y$
$- y = ln (\frac{x - 1}{5})$ $-y=ln((x-1)/5)$
$y = - ln (\frac{x - 1}{5})$ $y=-ln((x-1)/5)$ That is your answer!

Science

Math

Humanities

... and beyond

How do you find the inverse of 5/e^x+1?

1 Answer

Explanation:

Related questions

Impact of this question