# A step in an integral question multiplying logs and fractions

• May 30th 2011, 09:20 AM
dix
A step in an integral question multiplying logs and fractions
I have a problem understanding a part of a solution to a calculus question. It's a 'by-parts' question and it's close to the solution.

= [ln2 (2^4/4)-(2^5/16)] - [-1/16]

I don't understand this next step:
=4ln2 (2^4/4(4)) - (2^5/16) + (1/16)

I understand that this part in red is being multiplied by 4 in order to get 16s in the denominators, but understand is how it's being multiplied by 4. Are the log and fraction together? How does (2^4/4)x4 become (2^4/4(4))?

Thanks for any help :)
• May 30th 2011, 09:33 AM
e^(i*pi)
As it's written the fraction is not an argument of the log although for an integral output it seems very strange, what is the original question?

What you've wrote is $\displaystyle \ln(2) \times \dfrac{2^4}{4} - \dfrac{2^5}{16} - \left(-\dfrac{1}{16}\right)$ and when it reduces you get $\displaystyle 4\ln(2) - 2 + \dfrac{1}{16}$

All I can see what they've done in red is multiply that first term by $\displaystyle \dfrac{4}{4} = 1$

Could you post the original question so we can verify your solution?
• May 30th 2011, 10:06 AM
dix
Thnaks. Here is the question and answer:

http://i.imgur.com/7ezED.png
• May 30th 2011, 10:30 AM
e^(i*pi)
The solution to this integral is $\displaystyle [\dfrac{1}{16}x^4 \times (4\ln(x) - 1)]^2_1$ (I've tidied the syntax up a bit - post back if you didn't get this answer)

$\displaystyle [\dfrac{1}{16}(2^4) \times (4\ln(2)-1)] - [\dfrac{1}{16} \times -1] = 4\ln(2)-1 + \dfrac{1}{16} = \ln(16) - \dfrac{15}{16}$

I'm not sure how where you got the power of 5 from in your integral result from
• May 30th 2011, 11:18 AM
dix
Thanks again. You're right about the 5 in the integral. I've uploaded a pic of what I went through and highlighted the part I don't understand. The multiplying to get the 16 on the bottom just seems arbitrary and I can't make sense of it. http://i.imgur.com/vDIGn.jpg
• May 30th 2011, 11:39 AM
e^(i*pi)
For what it's worth the you have the letters the 'wrong' way round (according to convention). Your calculations are fine though

What's been done in the orange box is that they've multiplied top and bottom by 4, it's by no means an essential step and I'm not sure why they bothered to do it since it doesn't make a difference.

edit: it looks like they've done it to give a common denominator of 16 across the whole expression. Since that's been done we can factor out 1/16 from the fraction to give $\displaystyle \dfrac{1}{16} \left(2^4 \cdot 4 \ln(2) - 2^4 + 1\right) = \dfrac{1}{16}(64\ln(2) - 16 + 1)$

Looking at that first term (the one in orange) it is written as $\displaystyle \ln(2) \cdot \dfrac{2^4}{4}$. Since 2^4 = 16 and 16/4 = 4 we can cancel it down to $\displaystyle \ln(2) \cdot 4 = 4\ln(2)$. In turn this equals ln(16) from the log power law.

If we were to multiply top and bottom by 4 (as they did) we have : $\displaystyle \ln(2) \cdot \dfrac{2^4}{4} \cdot \dfrac{4}{4} = 4 \cdot \ln(2) \cdot \dfrac{2^4}{4 \cdot 4} = 4\ln(2) \cdot \dfrac{16}{16} = 4\ln(2)$

Does that make sense (I'm not too sharp at explaining)
• May 30th 2011, 12:09 PM
dix
Ah right yes, I can see it now. Thanks very much!