# Using logs in calculations

1. Nov 9, 2006

### John O' Meara

I wish to find the value of .8^(2/5) using logs. I can find the value of .8^(-2/5) as follows: =(log(8)X1/10)X-2/5
=(-1 + .9031)X-2/5 = (-.0969)X-2/5 = +.03876;
antilog(.03876) = 1.093;
Now to find .8^(2/5) my approach is the same:
log(.8)X2/5 = (log(8)X10^-1)X2/5
= (bar1 + .9031)X2/5 : what do I do next. (bar1 = -1)

2. Nov 9, 2006

### Integral

Staff Emeritus
It is hard to understand what you have done.

$$\log( .8^ \frac 2 5 )$$

$$= \frac 2 5 \log (.8)$$
$$= \frac 2 5 ( \log (8) - \log (10))$$

Excell tells me the answer should be ~.915

Last edited: Nov 9, 2006
3. Nov 9, 2006

### HallsofIvy

Staff Emeritus
John O'Meara's work looks exactly like what I used to do in highschool. (Of course, we did all calculations on an abacus back then!). Since a table of logarithms only gave logarithms for numbers between 1 and 10, write .8 as 8 x 10-1. Then log(.8)= log(8)- 1! It's hard to imagine anyone today doing it that way- a calculator will give immediately that log(.8)= -0.096910013008056414358783315826521, far more accurate than any table would be. 2/5 times that is
-0.038764005203222565743513326330608. (I got that, by the way, from the calculator supplied with Windows.)

Integral, log(.8) is negative. The value you give can't possibly be right.

4. Nov 10, 2006

### Integral

Staff Emeritus
Clarifcation:
My Excell value is for $.8 ^ \frac 2 5$ not the log.

5. Nov 11, 2006

### HallsofIvy

Staff Emeritus
Ah! Okay.

John O'Meara, after you have (-1+ .9031)X2/5 the obvious "next thing to do" is the multiplication: -2/5+ .36124= -.4+ .361234= -1+ .6+ .361234= -1+ .961234. Now look that up in the "body" of whatever log tables you are using: find the x that gives that logarithm. More simply you can use the calculator that comes with Windows to find the 'inverse' log of that: the inverse log of .961234 is 9.146056 so we have 9.146056x 10-1= 0.9146056. Actually, it is not at all difficult to use the Windows calculator to do .8.4 directly and see that that is, in fact, the correct answer.