What is the inverse of log^2_{3}(x-1) ?

The inverse of log^2_{3}(x-1) is 3^{sqrt(x)}+1,3^{-sqrt(x)}+1

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inverse of log^2_{3}(x-1)

inverse

lo g_{3}^{2} (x - 1)

3^{x} + 1, 3^{- x} + 1

Solution steps

y = lo g_{3} (x - 1)^{2}

Replace

x

with

y

x = lo g_{3} (y - 1)^{2}

Solve for

y, x = lo g_{3} (y - 1)^{2}

3^{x} + 1, 3^{- x} + 1

f (x) = (\frac{( x - 3 )}{8})^{\frac{1}{3}}

\frac{2 X}{X + 3}

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

5 x^{2} - \frac{6}{7}

g (x) = x^{2} + 2

What is the inverse of log^2_{3}(x-1) ?
The inverse of log^2_{3}(x-1) is 3^{sqrt(x)}+1,3^{-sqrt(x)}+1