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avatar+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? 

 Jan 24, 2018
 #1
avatar+8962 
+3

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......?   smiley

 Jan 24, 2018
edited by hectictar  Jan 24, 2018
 #2
avatar+119 
+2

Hm, ok thanks. Not sure what log means though...

Aleguan  Jan 24, 2018
 #3
avatar+8962 
+2

"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

 Jan 24, 2018

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