How do you find the inverse of  f(x) = ln(4 - 7x) + ln(-7 - 5x)?

Oct 23, 2016

There is no inverse, because the quadratic formula yields two equations that returns two values for any given x and an inverse must not do this.

Explanation:

Let $x = {f}^{-} 1 \left(x\right)$

Substitute ${f}^{-} 1 \left(x\right)$ everywhere there is an x:

$f \left({f}^{-} 1 \left(x\right)\right) = \ln \left(4 - 7 {f}^{-} 1 \left(x\right)\right) + \ln \left(- 7 - 5 {f}^{-} 1 \left(x\right)\right)$

Use the property of inverses $f \left({f}^{-} 1 \left(x\right)\right) = x$:

$x = \ln \left(4 - 7 {f}^{-} 1 \left(x\right)\right) + \ln \left(- 7 - 5 {f}^{-} 1 \left(x\right)\right)$

Use the identity $\ln \left(a\right) + \ln \left(b\right) = \ln \left(a b\right)$

x = ln((4 - 7f^-1(x)(-7 - 5f^-1(x)))

Use the inverse of the natural logarithm on both sides:

e^x = (4 - 7f^-1(x)(-7 - 5f^-1(x))

Use the F.O.I.L. method to perform the multiplication:

${e}^{x} = - 28 - 20 {f}^{-} 1 \left(x\right) + 49 {f}^{-} 1 \left(x\right) + 35 {\left({f}^{-} 1 \left(x\right)\right)}^{2}$

$35 {\left({f}^{-} 1 \left(x\right)\right)}^{2} + 29 {f}^{-} 1 \left(x\right) - 28 - {e}^{x} = 0$

At this point, we must declare that there is no inverse, because the quadratic formula yields two equations that returns two values for any given x and an inverse must not do this.

Let's proceed so that you can see the issue

Use the quadratic formula, $x = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$

${f}^{-} 1 \left(x\right) = \frac{- 29 + \sqrt{{29}^{2} - 4 \left(35\right) \left(- 28 - {e}^{x}\right)}}{2 \left(35\right)}$

AND

${f}^{-} 1 \left(x\right) = \frac{- 29 - \sqrt{{29}^{2} - 4 \left(35\right) \left(- 28 - {e}^{x}\right)}}{2 \left(35\right)}$

This is contrary to the purpose of an inverse function.

The purpose of an inverse function is, for any $y$ returned from the $f \left(x\right)$, give you back a single value of $x$ that created the $y$; this returns two values, therefore, no inverse.