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

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

inverse

y = \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 )}

10 - 2 p^{\frac{1}{2}}

f (x) = 4 + 3 x - 3

f (x) = \frac{3}{4} x + \frac{2}{7}

- 2^{x} + 3

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

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