What is the inverse of f(x)=ln(x^2-4)+pi,x<2 ?

The inverse of f(x)=ln(x^2-4)+pi,x<2 is sqrt(e^{x-pi)+4},-sqrt(e^{x-pi)+4}

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inverse of f(x)=ln(x^2-4)+pi,x<2

inverse

f (x) = ln (x^{2} - 4) + π, x < 2

e^{x - π} + 4, - e^{x - π} + 4

Solution steps

y = ln (x^{2} - 4) + π

Replace

x

with

y

x = ln (y^{2} - 4) + π

Solve for

y, x = ln (y^{2} - 4) + π

e^{x - π} + 4, - e^{x - π} + 4

\frac{5}{x} + 4

f (x) = 15 x^{2} + 2

f (x) = 3 4 - x + 3

f (x) = e^{\frac{( 5 - x ) \cdot l n ( 0.2 )}{5}}

f (x) = 15 x^{2} + 8

What is the inverse of f(x)=ln(x^2-4)+pi,x<2 ?
The inverse of f(x)=ln(x^2-4)+pi,x<2 is sqrt(e^{x-pi)+4},-sqrt(e^{x-pi)+4}