What is the inverse of #y= -log (1.05x+10^-2)# ?

1 Answer
Jun 25, 2017

Answer:

#f^-1(x)=(10^-x-10^-2)/1.05#

Explanation:

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.