What is the inverse of f(x)=1-log_{3}(x+1) ?

The inverse of f(x)=1-log_{3}(x+1) is 3^{1-x}-1

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inverse of f(x)=1-log_{3}(x+1)

inverse

f (x) = 1 - lo g_{3} (x + 1)

3^{1 - x} - 1

Solution steps

y = 1 - lo g_{3} (x + 1)

Replace

x

with

y

x = 1 - lo g_{3} (y + 1)

Solve for

y, x = 1 - lo g_{3} (y + 1)

3^{1 - x} - 1

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

- 1 + 4 x

lo g_{2} (x + 2)

f (x) = - 4 x^{7} - 1

f (x) = ln (\frac{e ^{x}}{e ^{x - 1}})

What is the inverse of f(x)=1-log_{3}(x+1) ?
The inverse of f(x)=1-log_{3}(x+1) is 3^{1-x}-1