What is the inverse of f(x)=sqrt(log_{1/2)(x-3)} ?

The inverse of f(x)=sqrt(log_{1/2)(x-3)} is (1/2)^{x^2}+3

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inverse of f(x)=sqrt(log_{1/2)(x-3)}

inverse

f (x) = lo g_{\frac{1}{2}} (x - 3)

(\frac{1}{2})^{x^{2}} + 3

Solution steps

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

Replace

x

with

y

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

Solve for

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

(\frac{1}{2})^{x^{2}} + 3

f (x) = \frac{9}{8 + x}

f (x) = arccos^{3} (\frac{3 x + 4}{x - 3})

f (x) = (x - 5)^{3} - 6

(x - 16)^{2}

f (x) = x^{2} - 4 x, x \geq 2

What is the inverse of f(x)=sqrt(log_{1/2)(x-3)} ?
The inverse of f(x)=sqrt(log_{1/2)(x-3)} is (1/2)^{x^2}+3