What is the domain of y = log [(4-t)^(2/3)] ?

[zacc]

Hello to all:

I was trying to solve the following question: what is the domain of the function y = log [(4-t)^(2/3)] for Real t. Here is my reasoning:

The domain of the log(x) is x>0.

I would have interpreted this question in terms of the log of the cubic root of (4-t)^2. Thus, I was thinking of the domain as the solution to the following inequality:

(4-t)^(2/3) > 0

The cubic root has the sign of the radicand and (4-t)^2 is positive for all t. The only restriction would be when 4-t=0 and this happens for t=4. Thus, my answer to the question is: the domain is ALL reals minus 4.

Apparently this is not the case as the answer to the problem is t<4. Now, I am suspecting this is because they are rewriting the function as: y=log[(2/3)*(4-t)] for which indeed the answer would be t<4.

So, I am really confused as to what is going on here. Do I always need to rewrite logarithmic functions to find their domains? Why? Please, help!

Thanks guys!

 

[Tanya Sharma]

Hi zacc...

Hello to all:

I was trying to solve the following question: what is the domain of the function y = log [(4-t)^(2/3)] for Real t. Here is my reasoning:

The domain of the log(x) is x>0.

I would have interpreted this question in terms of the log of the cubic root of (4-t)^2. Thus, I was thinking of the domain as the solution to the following inequality:

(4-t)^(2/3) > 0

The cubic root has the sign of the radicand and (4-t)^2 is positive for all t. The only restriction would be when 4-t=0 and this happens for t=4. Thus, my answer to the question is: the domain is ALL reals minus 4.

Apparently this is not the case as the answer to the problem is t<4. Now, I am suspecting this is because they are rewriting the function as: y=log[(2/3)*(4-t)] for which indeed the answer would be t<4.

So, I am really confused as to what is going on here. Do I always need to rewrite logarithmic functions to find their domains? Why? Please, help!

Thanks guys!

The item in red should be y=(2/3)log(4-t)

Its not about rewriting the logarithmic functions.Rather it is about fulfilling the properties of the function.

We know that log_baⁿ = nlog_ba .

Now,the domain of the given function needs to satisfy all the properties of logarithms .If you take domain to be R-{0} ,then clearly, the function is not defined for all the points of domain.More specifically ,the function is not defined for t≥4 .Hence,R-{0} cannot be the domain of the given function.

Whereas for t<4 ,the function is defined everywhere in the domain.

 

[Mark44]

Hello to all:

I was trying to solve the following question: what is the domain of the function y = log [(4-t)^(2/3)] for Real t. Here is my reasoning:

The domain of the log(x) is x>0.

I would have interpreted this question in terms of the log of the cubic root of (4-t)^2. Thus, I was thinking of the domain as the solution to the following inequality:

(4-t)^(2/3) > 0

The cubic root has the sign of the radicand and (4-t)^2 is positive for all t. The only restriction would be when 4-t=0 and this happens for t=4. Thus, my answer to the question is: the domain is ALL reals minus 4.

Apparently this is not the case as the answer to the problem is t<4. Now, I am suspecting this is because they are rewriting the function as: y=log[(2/3)*(4-t)] for which indeed the answer would be t<4.

No, it would be y = (2/3)log(4 - t), which is different from what you wrote, and has a domain of t < 4. Otherwise, your other work seems reasonable. If t = 4, (2/3)log[4 - t] and log[(4 - t)^2/3] are undefined, but if t > 4, log[(4 - t)^2/3] is defined, while (2/3)log(4 - t) is not.

Have you given us the complete problem? What's your reference that says that the domain is only t < 4?

So, I am really confused as to what is going on here. Do I always need to rewrite logarithmic functions to find their domains? Why? Please, help!

Thanks guys!

 

[zacc]

Tanya and Mark: Indeed I made a mistake and I should have written y=(2/3)*log(t-4). That was my intention. Sorry about that but it was quite late at night when I was writing. Thanks for the correction.

Hi zacc...



The item in red should be y=(2/3)log(4-t)

Its not about rewriting the logarithmic functions.Rather it is about fulfilling the properties of the function.

We know that log_baⁿ = nlog_ba .

Now,the domain of the given function needs to satisfy all the properties of logarithms .If you take domain to be R-{0} ,then clearly, the function is not defined for all the points of domain.More specifically ,the function is not defined for t≥4 .Hence,R-{0} cannot be the domain of the given function.

Whereas for t<4 ,the function is defined everywhere in the domain.

Hi Tanya: Thanks for the reply. I still don't see why my original reasoning is flawed. That is, (t-4)² is positive over all reals and so is its cubic root and thus its log should be defined But, please see below.

No, it would be y = (2/3)log(4 - t), which is different from what you wrote, and has a domain of t < 4. Otherwise, your other work seems reasonable. If t = 4, (2/3)log[4 - t] and log[(4 - t)^2/3] are undefined, but if t > 4, log[(4 - t)^2/3] is defined, while (2/3)log(4 - t) is not.

Have you given us the complete problem? What's your reference that says that the domain is only t < 4?

Mark: My only reference was asking the question to Wolframalpha: "what is the domain of the function y=log[(4-t)^(2/3)]" which returns t<4 as the answer. Requesting a plot of the function returns only the branch for t<4.

Tanya and Mark:

Now, I just realized that if I had asked Wolframalpha the question: what is the domain of the function y=log((t-4)^2) the answer in this case is all reals except 4. So, evidently the problem has to do with the way Wolframalpha handles the roots and maybe there is something that I am missing. If any of you have any insight it would be greatly appreciated. Thanks again.

 

[Mark44]

The answer you got is a result of how WA handles fractional exponents. I think this is what WA is doing:
$$(4 - t)^{2/3} = e^{ln(4 - t)^{2/3}} = e^{2/3 * ln(4 - t)}$$