#### HallsofIvy

Homework Helper
Byrgg said:
Please someone just explain how b ^ log_b (x) = x shows that log_b (x) and b ^ x are inverse. I know they're inverse, but I just need to know how you get this relationship out of the previously mentioned equation.
What DEFINITION of inverse are you using? In any book I've seen, two functions f(x) and g(x) are inverse to each other if and only f(g(x))= x and
g(f(x))= x.

#### Byrgg

Well the closest thing to that(the definition) that I found was something like log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

Sorry, but there's still some I'm not understanding here, how exactly does b ^ log_b (x) simlpify to x(proving the statement)? Could someone show me the actual step by step process of this?

Last edited:

#### d_leet

Byrgg said:
Well the closest thing to that(the definition) that I found was something like log_b (n) = x if and only if b ^ x = n, where b > 0, and b is not equal to one.

Sorry, but there's still some I'm not understanding here, how exactly does b ^ log_b (x) simlpify to x(proving the statement)? Could someone show me the actual step by step process of this?
The definition that you just stated defines the log function as the inverse to the exponential.

You had logb(n) = x iff bx = n.

So if you take the first equation there and raise each side to b then you have blogb(n) = bx. but we already said that bx = n so that means that blogb(n) = n.

#### Byrgg

So then if b ^ log_b (n) = b ^ x then that must mean log_b (n) = x. This is the same as writing b ^ x = n, and so if an exponential function took the form b ^ y = n then that's saying log_b (n) is the inverse of b ^ n?

Did I get that right? Is that why the equation says that log_b (x) and b ^ x are inverses?

#### Byrgg

Actually there's still something I'm not getting, on another forum someone is saying that when you consider that log_b (x) and b ^ x are inverses, there should be an immediate realization of the problem, anyone able to help me understand this exactly?

#### d_leet

Byrgg said:
Actually there's still something I'm not getting, on another forum someone is saying that when you consider that log_b (x) and b ^ x are inverses, there should be an immediate realization of the problem, anyone able to help me understand this exactly?
The definition of inverse functions has been posted several times already but I will do it again.

If two functions f(x) and g(x) are inverses of each other then

f(g(x)) = g(f(x)) = x

Do you understand this?

Thus since logb(x) and bx are inverse functions if we let f(x) = bx and g(x) = logb(x)

Then f(g(x)) = blogb(x)

And since we said that these functions are inverses then f(g(x)) = x which means that blogb(x) = x.

#### Byrgg

"Thus since log_b(x) and b ^ x are inverse functions if we let f(x) = b ^ x and g(x) = log_b(x)

Then f(g(x)) = b ^ log_b(x)"

How did you get from the first line to the second in this? I'm sorry, I'm just not seeing how you did it.

I know that f(g(x)) = g(f(x)) = x, but I got confused when you did the above...

#### d_leet

Byrgg said:
"Thus since log_b(x) and b ^ x are inverse functions if we let f(x) = b ^ x and g(x) = log_b(x)

Then f(g(x)) = b ^ log_b(x)"

How did you get from the first line to the second in this? I'm sorry, I'm just not seeing how you did it.

I know that f(g(x)) = g(f(x)) = x, but I got confused when you did the above...

The log function is DEFINED as the inverse of the exponential function.

If we have a function y=f(x)

then there is a function x = g(y), such that the function y=g(x) is the inverse of f

Now let's look at this in terms of the exponential and logarithmic functions.

we have y= bx
Now we want to define a function x = g(y) that will be the inverse of this function, and we define this function to be the logarithm so that if

y = bx
then
x = logb(y) these two equations are the exact same thing, and so y = logb(x) is the inverse of y = bx.

Again letting f(x) = bx
and g(x) = logb(x)

so when I take f(g(x)) all that I am doing is taking teh original x in f(x) and putting g(x) there instead

f(x) = bx
f(g(x)) = bg(x)
f(g(x)) = blogb(x) = x

Is this helping at all?

#### Byrgg

Ok, I think I've got it now:

let f(x) = b ^ x and g(x) = log_b(x)

f(x) = b ^ x

and x = f(g(x))

therefore subbing all the x's with g(x) brings the folllowing about:

f(g(x)) = b ^ g(x) = x
but g(x) = log_b(x)
therefore f(g(x)) = b ^ log_b(x) = x

Yes I think I finally got it all, did that look right? It's basically just what you did(I'm pretty sure) but I sort of put it into my own words.

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