What f-1(x) of f(x)=(1/3)^x

You want the inverse of:

y = (1/3)^x ?????
hmmm maybe try
x = (1/3)^y
ln x = y ln(1/3)

y = ln x /ln(1/3) = -ln x ln 3

y = - 1.1 ln x approximately

or, to avoid introducing a new base,

f^-1(x) = log1/3x = -log3x

log and exponentiation are inverses, just like

+ and -
* and /
square root and square

We just don't have any special symbols to denote the operation.

To find f-1(x) of f(x)=(1/3)^x, we need to find the inverse function.

To begin, let's rewrite the function with the inverse notation:
x = (1/3)^f-1(x)

The first step is to solve for f-1(x) by isolating it. We can do this by taking the logarithm base (1/3) on both sides of the equation:

log_base(1/3)(x) = f-1(x)

Since the base is (1/3), we can use the logarithmic property that states: log_base(a)(a^b) = b. Applying this property to the equation, we get:

f-1(x) = log_base(1/3)(x)

Therefore, the inverse function is f-1(x) = log_base(1/3)(x).