+119 Just a little quickie, since Google won't help me out.
g(x) = 9^(x)-3
h(x) = g^(-1)(x)
Is g^(-1)(x) equal to (1/g)(x) or 1/(g(x))? In other words, would g^(-1) be equivalent to 1 over all of 9^(x)-3?
+9481 In this context g-1(x) does not mean g(x) raised to the power of negative 1 .
It means the inverse function of g(x) . We can find g-1(x) like this....
g(x) = 9^x - 3
Instead of g(x) , write y .
y = 9^x - 3
Now we want to solve this equation for x .
y + 3 = 9^x
log(y + 3) = log(9^x)
log(y + 3) = x log 9
log(y + 3) / log 9 = x
To get the inverse function, swap x and y .
log(x + 3) / log 9 = y This is the inverse function. We can say...
g-1(x) = log(x + 3) / log 9
So... g-1(x) does not mean 1 / g(x) .
Neither does sin-1x = (sin x)-1 , even though sin2x = (sin x)2
This is something that has bothered me before. Maybe there is a reason why it is okay and I just don't understand it......? 
+9481 "log" is short for "logarithm" .
This can probably explain better than I can: https://youtu.be/Z5myJ8dg_rM
Here's a simpler example of finding the inverse function:
f(x) = x + 1
Instead of f(x) , let's say y .
y = x + 1 Solve for x .
y - 1 = x
Swap x and y .
x - 1 = y This is the inverse.
f-1(x) = x - 1
