J jeca86 Junior Member Joined Sep 9, 2005 Messages 62 Sep 25, 2005 #1 Find the inverse equation of f(x)= ln(x+ (square root of (x^2 +1))). My answer came out to be f^-1 (x)= 2lnx. I'm not sure if it's right.
Find the inverse equation of f(x)= ln(x+ (square root of (x^2 +1))). My answer came out to be f^-1 (x)= 2lnx. I'm not sure if it's right.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Sep 25, 2005 #2 f(x)=y= ln(x+ √(x<SUP>2</SUP>+1)). x=ln(y+ √(y<SUP>2</SUP>+1)). e<SUP>x</SUP>=y+ √(y<SUP>2</SUP>+1). Now someway you must solve for y.
f(x)=y= ln(x+ √(x<SUP>2</SUP>+1)). x=ln(y+ √(y<SUP>2</SUP>+1)). e<SUP>x</SUP>=y+ √(y<SUP>2</SUP>+1). Now someway you must solve for y.
