What is the inverse of 4/(ln(x^2-1)) ?

The inverse of 4/(ln(x^2-1)) is sqrt(e^{4/x)+1},-sqrt(e^{4/x)+1}

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inverse of 4/(ln(x^2-1))

inverse

\frac{4}{ln ( x ^{2} - 1 )}

e^{\frac{4}{x}} + 1, - e^{\frac{4}{x}} + 1

Solution steps

y = \frac{4}{ln ( x ^{2} - 1 )}

Replace

x

with

y

x = \frac{4}{ln ( y ^{2} - 1 )}

Solve for

y, x = \frac{4}{ln ( y ^{2} - 1 )}

e^{\frac{4}{x}} + 1, - e^{\frac{4}{x}} + 1

f (x) = x + 30

f (x) = \frac{6 - x}{3}

y = \frac{5}{x - 3} + 6

f (x) = x^{2} - 20

f (x) = \frac{x + 2}{x + 11}

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