What is the inverse of f(x)=1+log_{1/e}(2-x) ?

The inverse of f(x)=1+log_{1/e}(2-x) is -(1/e)^{x-1}+2

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inverse of f(x)=1+log_{1/e}(2-x)

inverse

f (x) = 1 + lo g_{\frac{1}{e}} (2 - x)

- (\frac{1}{e})^{x - 1} + 2

Solution steps

y = 1 + lo g_{\frac{1}{e}} (2 - x)

Replace

x

with

y

x = 1 + lo g_{\frac{1}{e}} (2 - y)

Solve for

y, x = 1 + lo g_{\frac{1}{e}} (2 - y)

- (\frac{1}{e})^{x - 1} + 2

f (x) = x 3 + 3 x 2 + 1

f (x) = \frac{( 7 x + 3 )}{( x - 5 )}

f (x) = d + 1

\frac{- 3 x + 4}{7 x + 9}

ln (4 x + 2)

What is the inverse of f(x)=1+log_{1/e}(2-x) ?
The inverse of f(x)=1+log_{1/e}(2-x) is -(1/e)^{x-1}+2