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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{3 x}{x - 9}

h (x) = \frac{9}{x + 11}

f (x) = 2 (x - 5)^{2} - 6

f (x) = 15 + 3 x

f (x) = x^{2} - 1, x \geq 1

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