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

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

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

inverse

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

2^{x} - 1, - 2^{x} - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

2^{x} - 1, - 2^{x} - 1

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