What is the range and domain of a logarithm function?

[Byrgg]

Quick question about logarithms

I'm only in gr.11 right now, and I was looking at a gr.12 textbook, since this isn't really homework I decided to post this here, is that ok or is this the wrong board?

Well anyway, it asked what the range of of the function y = b ^ n would be (and then what that means for the domain of log(b) n) and also if log(b) n could ever be negative or 0.

Could someone please help?

Oh and an explained answer(not just a one word one) is hoped for.

 

[HallsofIvy]

You can't say what the range is until you give the domain! Also, it is of importance whether b is less than or larger than 0 (or equal to 0!).

I would think, from your use of "n" rather than "x", that the domain is all positive integers but you question makes more sense if the domain is all real numbers.

 

[Byrgg]

I'm gr.11, and I'm looking in a gr.12 textbook, I read about logarithms and a few questions asked the following:

What is the range of the function y = b ^ n? What does this tell you about the domain of log(b)n? Explain.

Could the value of log(b)n ever be a negative number? Explain.
Could the value of log(b)n ever be zero? Explain.

It's not really homework, but this is straight out of a textbook so put it here. I don't really know what to try and figure out, so I really want some help with this.

Oh, and I can't get the imaging thing to work, so the b in the brackets is subscript just to let you know.

 

[Byrgg]

Err, I'm not sure of all the details, that's what's in the textbook. There's no answer you can give?

Word for word here is what it says:

What is the range of the function y = b ^ n? What does this tell you about the domain of log(b) n? Explain.

Could the value of log(b) n ever be a negative number? Explain.
Could the value of log(b) n ever be zero? Explain.

The b's in the brackets are subscript by the way, I can't seem to get the imaging thing to work.

 

[gnomedt]

I suppose it means for different values of b and n, what kind of numbers can you get out of $$b^n$$? I suggest you experiment with this problem, for positive and negative b and n, which (with aid of a calculator).

Are you familiar with inverse functions and the logarithm? The question involves some simple properties of inverse functions.

 

[mathwonk]

assuming the variable is n, and b >0, then a;; values of b^n are positive, and if n can take any real value, then all positive reals can occur as valkues of b^n.consequently, log can nevre be negative, but it sounds as if you need a better book. try courant calculus.

 

[mathwonk]

wqell for some reason i cannot edit my stupid answer to the previous question. i.e. i should have said log is only defined for positive numbers, but can have as values any real number. i.e. the inpuit of exp is any real and the output is positive, so vice versa, the input of log is positive numbers and the output is all reals.

 

[Byrgg]

If you input negative or zero numbers could the output be negative or zero?

 

[Byrgg]

"Are you familiar with inverse functions and the logarithm? The question involves some simple properties of inverse functions."

Oh and what do you mean by properties of inverse functions?

 

[arildno]

If you input negative or zero numbers could the output be negative or zero?

You cannot have a negative real number (or zero) as input in the log function and get out a real number. That is, for the real log function, its maximal domain is the set of strictly positive real numbers.

 

[Byrgg]

Why can't you put a negative or zero number into a log function and get out a real one what variable in the formula exactly do you test that with say the? Also, how would I do something like this? b ^ log(b)x = 1, it asks to prove that the statement is true and I don't know what to with a log in the exponent.
Once again, the b in the brackets is subscript.

 

[Byrgg]

Someone please help.

 

[HallsofIvy]

Please help? Have you understood what has been said so far?

If b is positive and the domain is all real numbers, then:

a) bⁿ, where n is a positive integer, is just b multiplied by itself n times: a positive number.
b) b⁰= 1, a positive number.
c) b^-n, where n is a positive integer so -n is a negative integer, is 1/bⁿ and 1 over a positive real number is a positive real number.
d) b^m/n, where m is an integer and n is a positive integer (so m/n is any rational number) is $\^n\sqrt{b^m}$ and $\^n\sqrt{ }$ is defined as the positive root.
e) if x is irrational, it's much more "technical": there exist a sequence of rational numbers, {r_n} that converge to x. b^x is the limit of the sequence $\left{b^{r_n}\right}$. The limit of a sequence of positive numbers cannot be negative and, in this case, cannot be 0.

The range of the function b^x, for b> 0, is the set of all positive real numbers. Since "inverting" a function, in this case going from b^x to log_b(x), reverses domain and range, the domain of log_b(x) is the set of all positive real numbers.

PLEASE read your problems more carefully. You are NOT asked to prove that $b^{log_b(x)}= 1$ because it isn't true. What you are asked to prove is that $b^{log_b(x)}= x$, which is precisely saying that b^x and log_b(x) are inverse functions. HOW you would prove that depends on what definitions of log_b(x) and b^x you are using.

 

[Byrgg]

Oh woops sorry, yeah I did read the question wrong. Can't b be a negative number though(unrelated to the misread question)?

 

[VietDao29]

Ok, I'd suggest you to re-read the book. Yes, reading the book before having a lecture is a very very good way to learn new things. You can first try to read the book, and understand what it says. Then, you can try to think a little bit further (making up some of your own questions 'why?', 'how?'... and try to answer them), then try to answer the problems the books provide to understand the concept better.
Ok, so first, let's begin with the function: a^x. This function is defined for all real x's, and for a > 0. But can a be less than 0? Why? You can think about the case when x is irrational, or some rational values of x like x = 0.5, 0.25, ...
And it's range is $$] 0, \ \infty [$$. You can graph some functions to see.
Now, read the book again to see if you can answer all the questions. :)

 

[VietDao29]

Oh woops sorry, yeah I did read the question wrong. Can't b be a negative number though(unrelated to the misread question)?

Since you are still in the reals, b cannot be negative. Why? Just think about the irrational value of x (i.e some values that cannot be expressed as any fraction m / n, where m, n are integers, e.g $$\sqrt{2}$$). Or some rational value of x like x = 0.5, x = 0.25, ...
Is $$\sqrt{-2}$$, or $$(-2) ^ {\sqrt{2}}$$ defined in the reals? :)
You can try to re-read the book again. If you find it too hard to understand, you can give mathwonk's recommendation a try. :)

 

[dav2008]

You already posted this question and received numerous replies: https://www.physicsforums.com/showthread.php?t=123695

Try to keep your questions to one forum.

 

[Byrgg]

Ok, so b is always a real number? Is there an explanation as to why it's treated this way, or is that just how you use it in logarithms?


"HOW you would prove that depends on what definitions of logb(x) and bx you are using."

Definitions of them? What do you mean? There isn't just a way to go about proving it?

 

[Byrgg]

And also, how do you know that b ^ log(b) x = x is saying that b ^ x and log(b) x are inverse functions? Not that I'm doubting this, I'm just curious as to how you go about figuring that out.

Oh and I read about how b in the exponential functions is > 0 in the text somewhere as well so I'm not really confused about that anymore. The above problem is the only thing confusing me right now.

 

[gnomedt]

It might make more sense for you if you looked as log_b(x) as the inverse function to exp_b(x) = b^x (where underscore means subscript). If f(x) and g(x) are inverse functions, then if y=f(x), then by definition x=g(y), and then combining those two equations you get y=f(g(y)) or x=g(f(x)). So if log_b(x) is the inverse function to exp_b(x), then it must be that x = exp_b(log_b(x)) and x = log_b(exp_b(x)).

(You never see it written exp_b(x) though. I just made that up.)

I suggest you read more about the logarithm, these are very basic questions that Google can answer better than we can (and with less sass).

 

[Byrgg]

Ok I think I've figured out that question. An inverse of an exponential function would reverse the spots of x and y where y = b ^ x right?
The inverse would be x = b ^ y or simply log_b (x) correct?

If b ^ log_b (x) = x, then the inverse would reverse the exponent with the number on the other side of the equation:

b ^ x = log_b (x) ^ -1
the inverse of log_b (x) is b ^ x right?

Then you just write as b ^ x = b ^ x

Someone please tell me I did that right.

 

[HallsofIvy]

I have merged all posts to this question in the "general math" section to this.

 

[Byrgg]

It was said earlier that the question stating b ^ log_b (x) = x is precisely saying that b ^ x and log_b (x) are inverses. How do you know that's what it's saying? I'm guessing because if you reverse x and log_b (x) in that statement, you get the equation which you must prove. By reversing them to the previous statement(b ^ x and log_b (x)), this obviously means that the equation is no longer equal, does reversing the two mean that they are inverses?

Someone please tell me if I'm right.

 

[Byrgg]

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.

 

[d_leet]

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.

Suppose we have 2 functions f(x) and g(x), f and g are inverses of each other if
f(g(x)) = g(f(x)) = x, you can think of it as one function undoing the other to bring you back to what you started with which was x, in other words if you do something to x by inputting it into f(x) you can undo that by putting f(x) into g(x) and you will get x back, similarly if you starting with g(x) and putting that into f(x).

Now let's suppose that f(x) = b^x
and g(x) = log_b(x)

then if these two functions are inverses f(g(x)) = x
and f(g(x)) = b^log_b(x) which does equal x because the log function is defined as the inverse to the exponential function.

That is why b^log_b(x) = x shows that these functions are inverses just as log_b(b^x) = x shows this same relationship.

 

[HallsofIvy]

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?

 

[d_leet]

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 log_b(n) = x iff b^x = n.

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

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 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)

And since we said that these functions are inverses then f(g(x)) = x which means that b^log_b(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]

"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= b^x
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 = b^x
then
x = log_b(y) these two equations are the exact same thing, and so y = log_b(x) is the inverse of y = b^x.

Again letting f(x) = b^x
and g(x) = log_b(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) = b^x
f(g(x)) = b^g(x)
f(g(x)) = b^log_b(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.