[tex]Domain:\\x+1 > 0\to x > -1\\\\f(x)=\log_5(x+1)\to y=\log_5(x+1)\\\\\text{exchange x to y}\\\\\log_5(y+1)=x\qquad\text{solve for y}\\\\\log_5(y+1)=x\iff y+1=5^x\qquad\text{subtract 1 from both sides}\\\\y=5^x-1\\\\f^{-1}(x)=5^x-1\ for\ x > -1\\\\f^{-1}(2)=5^2-1=25-1=24\\\\Answer:\ \boxed{f^{-1}(2)=24}[/tex]
