What is the inverse of f(x)=log_{10}(1-x^2) ?

The inverse of f(x)=log_{10}(1-x^2) is sqrt(-10^x+1),-sqrt(-10^x+1)

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inverse of f(x)=log_{10}(1-x^2)

inverse

f (x) = lo g_{10} (1 - x^{2})

- 1 0^{x} + 1, - - 1 0^{x} + 1

Solution steps

y = lo g_{10} (1 - x^{2})

Replace

x

with

y

x = lo g_{10} (1 - y^{2})

Solve for

y, x = lo g_{10} (1 - y^{2})

- 1 0^{x} + 1, - - 1 0^{x} + 1

\frac{( 4 x - 1 )}{2 x + 3}

y = 2 x^{2} - 1

f (x) = \frac{3 x - 10}{2 x}

\frac{x + 4}{x + 7}

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

What is the inverse of f(x)=log_{10}(1-x^2) ?
The inverse of f(x)=log_{10}(1-x^2) is sqrt(-10^x+1),-sqrt(-10^x+1)