What is the inverse of f(x)=(b^x+b^{-x})/2 ?

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

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inverse of f(x)=(b^x+b^{-x})/2

inverse

f (x) = \frac{b ^{x} + b ^{- x}}{2}

\frac{ln ( x + x ^{2} - 1 )}{ln ( b )}, \frac{ln ( x - x ^{2} - 1 )}{ln ( b )}

Solution steps

y = \frac{b ^{x} + b ^{- x}}{2}

Replace

x

with

y

x = \frac{b ^{y} + b ^{- y}}{2}

Solve for

y, x = \frac{b ^{y} + b ^{- y}}{2}

\frac{ln ( x + x ^{2} - 1 )}{ln ( b )}, \frac{ln ( x - x ^{2} - 1 )}{ln ( b )}

f (x) = 2 - 5 - x

1 + e^{- t}

- 2 x - 4

f (x) = 11 + 6 x - 6

x + 4 + 1

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