What is the inverse of f(x)=log_{2}(4x^2-1) ?

The inverse of f(x)=log_{2}(4x^2-1) is (sqrt(2^x+1))/2 ,-(sqrt(2^x+1))/2

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inverse of f(x)=log_{2}(4x^2-1)

inverse

f (x) = lo g_{2} (4 x^{2} - 1)

\frac{2 ^{x} + 1}{2}, - \frac{2 ^{x} + 1}{2}

Solution steps

y = lo g_{2} (4 x^{2} - 1)

Replace

x

with

y

x = lo g_{2} (4 y^{2} - 1)

Solve for

y, x = lo g_{2} (4 y^{2} - 1)

\frac{2 ^{x} + 1}{2}, - \frac{2 ^{x} + 1}{2}

2 (x - 2)^{2} = 8 (7 + y)

f (x) = \frac{4 + x}{x}

f (x) = \frac{3 x}{5} + 1

f (x) = - \frac{3}{4} x + \frac{5}{2}

f (x) = \frac{5 x + 3}{2 x - 1}

What is the inverse of f(x)=log_{2}(4x^2-1) ?
The inverse of f(x)=log_{2}(4x^2-1) is (sqrt(2^x+1))/2 ,-(sqrt(2^x+1))/2