# Really Really Quick Questions!

I just got my calculus midterm back, and I lost one mark on an area problem. I did the problem properly, but I may have made a mistake at the end.

I end up with arctan1-1/2ln3-1/2ln1 (this step did not have an "x")
I simplified this to pi/4-1/2ln3 (here, the grader put a 2 over my 3 in ln3)
then I made it pi/4-ln(3^1/2) (there was a question mark around this)

so my quick questions are:

why was there are 2 over my three? (I thought 1/2ln1 was equal to zero.)
why was there a question mark around the last part (I thought alnx was equal to ln(x^a)

Hopefully this makes sense, post and I can clarify

thanks SO much in advance if you helped!

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cristo
Staff Emeritus
It appears to me that the marker is unsure on properties of logarithms--you are correct.

The correct properties are: loga-logb=log(a/b) (1) and alogb=log(ba).

I end up with arctan1-1/2ln3-1/2ln1 (this step did not have an "x")
I simplified this to pi/4-1/2ln3 (here, the grader put a 2 over my 3 in ln3)
I can only think the marker has thought that ln3-ln1=ln(3-1)=2.* This is not correct. Of course, ln1=0. You can simplify the expression using (1) to give ln3-ln1=ln(3/1)=ln3.
then I made it pi/4-ln(3^1/2) (there was a question mark around this)
Again, this is correct.
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edit:*In retrospect, I don't see where the marker got ln2 from, since the expression would be -1/2{ln3+ln1} which, would become -1/2{ln4} using the incorrect rule. Anyway, the fact remains that you are correct.

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Ok thanks! That's what I thought...hopefully I'll be able to get my prof to change my mark!