Given: #f(x) = -log (1.05x+10^-2) #
Let #x = f^-1(x)#
#f(f^-1(x)) = -log (1.05f^-1(x)+10^-2) #
By definition #f(f^-1(x)) = x#
#x = -log (1.05f^-1(x)+10^-2) #
Multiply both sides by -1:
#-x = log (1.05f^-1(x)+10^-2) #
Make both sides the exponent of 10:
#10^-x = 10^(log (1.05f^-1(x)+10^-2))#
Because 10 and log are inverses, the right side reduces to the argument:
#10^-x = 1.05f^-1(x)+10^-2#
Flip the equation:
#1.05f^-1(x)+10^-2=10^-x#
Subtract 10^-2 from both sides:
#1.05f^-1(x)=10^-x-10^-2#
Divide both sides by 1.05:
#f^-1(x)=(10^-x-10^-2)/1.05#
Check:
#f(f^-1(x)) = -log (1.05((10^-x-10^-2)/1.05)+10^-2)#
#f(f^-1(x)) = -log (10^-x-10^-2+10^-2)#
#f(f^-1(x)) = -log (10^-x)#
#f(f^-1(x)) = -(-x)#
#f(f^-1(x)) = x#
#f^-1(f(x)) =(10^-(-log (1.05x+10^-2) )-10^-2)/1.05#
#f^-1(f(x)) =(10^(log (1.05x+10^-2) )-10^-2)/1.05#
#f^-1(f(x)) =(1.05x+10^-2-10^-2)/1.05#
#f^-1(f(x)) =(1.05x)/1.05#
#f^-1(f(x)) =x#
Both conditions check.
