What is the inverse of (2^x-2^{-x})/2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (2^x-2^{-x})/2

inverse

\frac{2 ^{x} - 2 ^{- x}}{2}

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 3 (x - 1)^{2} + 2

ln (5 x + 3)

f (x) = 10 x - 4

f (x) = \frac{4 x ^{3}}{5} - \frac{7}{6}

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