# Matural log

1. Jul 28, 2008

### Brunno

Calcule: $$ln\frac{273}{263}$$

2. Jul 28, 2008

### Defennder

That doesn't make any sense. What are we supposed to evaluate it to? Isn't it already given in the simplest form?

3. Jul 28, 2008

### Brunno

Re: Natural log

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

$$ln3*91$$
$$ln3+ln9,1*10$$
$$ln3+ln9,1+ln10$$

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

Is it right?

4. Jul 28, 2008

### symbolipoint

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.

5. Jul 28, 2008

### Brunno

6. Jul 28, 2008

### HallsofIvy

Staff Emeritus
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)?

7. Jul 28, 2008

### Brunno

$$ln263=ln2+ln1,315+ln10+ln10$$
We have that $$ln2=0.7;ln1,3=0.26;ln10=2.3$$ The sum is equable to 5.56.

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

I think that's the corect way isn't it?

8. Jul 29, 2008

### maze

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:

9. Jul 29, 2008

### HallsofIvy

Staff Emeritus
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.

10. Jul 29, 2008