Any ideas why this law of logs problem is marked incorrect?

[Niaboc67]

Thanks

 

[ehild]

Thanks

It is correct, but you can expand x²-25 further. And note that |x|>5

 

[Niaboc67]

so instead of 25 it would be 5?

 

[Mark44]

It is correct, but you can expand x²-25 further. And note that |x|>5

What ehild means in the last sentence is that you must have |x| > 5.

so instead of 25 it would be 5?

No, what he's saying is that you can factor x² - 25.

 

[ehild]

I also meant that the expression is not defined for |x| ≤ 5. You have to exclude it. And when expanding x²-25, log(x-5) is only defined if x>5,

 

[Fredrik]

The original expression is defined for all $x$ such that $|x|>5$. (These values of $x$ make the thing under the square root positive). But $\log(x-5)$ is defined for all $x$ such that $x>5$. So $\log(x-5)$ isn't defined for all $x$ such that the original expression makes sense. This seems like a good reason to not do the rewrite $\log(x^2-25)=\log(x+5)+\log(x-5)$.

 

[ehild]

The original expression is defined for all $x$ such that $|x|>5$. (These values of $x$ make the thing under the square root positive). But $\log(x-5)$ is defined for all $x$ such that $x>5$. So $\log(x-5)$ isn't defined for all $x$ such that the original expression makes sense. This seems like a good reason to not do the rewrite $\log(x^2-25)=\log(x+5)+\log(x-5)$.

Yes, the expression can not be really expanded further for all x. Why was it marked incorrect then?

 

[BruceW]

I'm guessing it is an online system which the answer is typed into, and automatically marked... So maybe it is because the 5 was placed right before the log, as in
5log, and maybe the computer did not recognize this as 5*log ?

 

[Niaboc67]

I understand now. The x-25 could have been factored more. Thanks guys!

 

[Mark44]

I understand now. The x-25 could have been factored more. Thanks guys!

Just to be clear, that should be x² - 25.

 

[Fredrik]

I understand now. The x-25 could have been factored more. Thanks guys!

The $x^2-25$ can be factored, but we also said that it shouldn't be, because you don't want a final answer that makes sense for a smaller set of values of $x$ than the original expression. For example, the original expression makes sense when $x=-7$, but an expression that contains $\log(x-5)$ doesn't. So we don't know why the answer you posted was marked incorrect. See BruceW's post for a possible reason.