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Homework Help: Matural log

  1. Jul 28, 2008 #1
    Calcule: [tex]ln\frac{273}{263}[/tex]
  2. jcsd
  3. Jul 28, 2008 #2


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    That doesn't make any sense. What are we supposed to evaluate it to? Isn't it already given in the simplest form?
  4. Jul 28, 2008 #3
    Re: Natural log

    Is it?I'm sorry I don't quite understand this...Well,could this be the answer:


    According to the table of natural logs:[tex]ln3=1.09;ln9,1=2.2;ln10=2.3[/tex]
    So:[tex]ln= 1.1+2.2+2.3=5.6[/tex]

    Is it right?
  5. Jul 28, 2008 #4


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    The first question to ask is, can you reduce the rational number? The quantities are not multiples of 3 or 9. Anything else possible? If not, then either use a calculator for natural logarithm of that ratio or look in tables for log_e of the numerator minus log_e of the denominator.
  6. Jul 28, 2008 #5
    So,It's correct my answer,right?
  7. Jul 28, 2008 #6


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    You still haven't made clear what the question is!

    Yes, 271= 3*91= 3*9.1*10 so ln(271)= ln(3)+ ln(9.1)+ ln(10). and, to one decimal place, ln(271)= 5.6. But what does that have to do with ln(271/263)?
  8. Jul 28, 2008 #7
    We have that [tex]ln2=0.7;ln1,3=0.26;ln10=2.3[/tex] The sum is equable to 5.56.

    At that time I was looking for for [tex]ln\frac{273}{263}[/tex],that's equal to [tex]ln273-ln263[/tex].It answer is,by a calculator,roughly 0.0373177.

    I think that's the corect way isn't it?
  9. Jul 29, 2008 #8
    Since 263/273 is only slightly larger than 1, you could use the taylor series expansion for x near 1. To the first power ln(1+a) ~ a, to the second ln(1+a) ~ a - 1/2a^2. These give results of .03802 and .03729 respectively. The correct answer is .03732.

    Or do this:
  10. Jul 29, 2008 #9


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    Or just use your calculator to find "LN (273/263)". I still don't understand what the problem is. If the problem is just to find an (approximate) value, why break it into parts? Why not just do it directly- find 273/263 and then find the logarithm of that.
  11. Jul 29, 2008 #10
    Thankyou both guys,You were very helpful.I already got my question answered.Was just that simple!
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