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.
