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

• Niaboc67

#### Niaboc67

Thanks

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

so instead of 25 it would be 5?

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.

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,

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)##.

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?

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 ?

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

I understand now. The x-25 could have been factored more. Thanks guys!
Just to be clear, that should be x2 - 25.

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.