How do you find the inverse of $f (x) = ln (3 - 2 x) + 3$ $f(x)=ln (3-2x)+3$ ?

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Answer 1 · 2015-11-22T22:04:02

Inverse is $f^{- 1} (x) = \frac{3}{2} - \frac{e^{x - 3}}{2}$ $f^-1(x) = 3/2 - (e^(x-3))/2$

Make $x$ $x$ the subject in this equation $y = ln (3 - 2 x) + 3$ $y= ln(3-2x) + 3$

Step 1) $y - 3 = ln (3 - 2 x)$ $y-3 = ln(3-2x)$

Step 2) Make exponential expression on both sides to get rid of $ln$ $ln$ so $e^{y - 3} = 3 - 2 x$ $e^(y-3) = 3-2x$

Step 3) The final form $x = \frac{3}{2} - \frac{e^{y - 3}}{2}$ $x= 3/2 - (e^(y-3))/2$

Step 4) Final step replace x by $f^{- 1} (x)$ $f^-1 (x)$ and $y$ $y$ by $x$ $x$ .

Hope that helps,
Cheers.

How do you find the inverse of f(x)=ln (3-2x)+3f(x)=ln(3−2x)+3f(x)=ln (3-2x)+3?