How do you find the inverse of #y = -(1/3)^x #?

Given: #f(x) = -(1/3)^x; -oo < x < oo#

Here is the graph of the function:

Please notice that the range of the function is #-oo < y < 0#; this will be the domain for the inverse: #-oo < x < 0#

Substitute #f^-1(x)# for every x:

#f(f^-1(x)) = -(1/3)^(f^-1(x)); -oo < x < 0#

The left side becomes x, because it is the definition of an inverse that #f(f^-1(x)) = x = f^-1(f(x))#:

#x = -(1/3)^(f^-1(x)); -oo < x < 0#

Multiply both sides by -1:

#-x = (1/3)^(f^-1(x)); -oo < x < 0#

Because x can only be negative, we can use a logarithm of undetermined base, b, on both sides:

#log_b(-x) = log_b((1/3)^(f^-1(x))); -oo < x < 0#

Use the property of logarithms #log_b(a^c) = (c)log_b(a#

#log_b(-x) = (f^-1(x))log_b(1/3); -oo < x < 0#

Use the property of logarithms #-log_b(a) = log_b(1/a)#

#log_b(-x) = (f^-1(x))(-log_b(3)); -oo < x < 0#

Divide both side by #(-log_b(3))#:

#f^-1(x) = -log_b(-x)/log_b(3); -oo < x < 0#

You can use base 10, base e, or any base that you choose.

I will leave it to you to show that #f(f^-1(x)) = x and f^-1(f(x)) = x#. You should always check this.