Solve 37 x 0.785 w/ Logarithms (Base 10, e & 5)

[chemical]

im sick of logs

use base and index steps to show the answer to 37 x 0.785 using logarithms with a base of 10, e and 5.

i just want to know how to start this problem.

thanks in advance

 

[HallsofIvy]

I'm really not sure what "base" and "index" steps are.

And I must admit that I'm surprised to see anyone using logarithms to actually do multiplications in this era! It's an interesting historical fact, but I consider the use of logarithms as the "inverse" of exponentials to be a much more important use.

I suspect what you are looking at is:

The log(37 x 0.785)= log(37)+ log(0.785) in any base. If you were doing this in base 10 and looking the values up in a table (why would anyone do that anymore), you would find that the table only contains logarithms of numbers from 1 to 10 and so does not contain either
37 or 0.785. What you would do is write 37 as 3.7* 10 and then look up the logarithm of 3.7 (it's 0.5682 to 4 decimal place- I used a calculator, not a table!) and note that the log of 10 is 1 simply because 10= 10¹ so that the log of 37 is 1.5682. Now write 0.785 as 7.85*10^-1. Look up the log of 7.85 (it's 0.8949-guess how I got that!) and of course, the log of 10^-1 is -1. In the "olden days" we would simply write that as
0.8949-1. Now, we know that log(37*0.785)= 1.5682+ 0.8949- 1= 1.4631 (yes, I used a calculator to do that sum!). Now "looking up" the anti-logarithm of that (once upon a time you just used your table of logarithms in reverse), the anti-log of 0.4631 is 2.904. Since the anitlogarithm of 1 is 10, the result is 2.904*10= 29.04 (remarkably enough, that's what my calculator gives as the product of 37 and 0.785!).

As far as base e and base 5 are concerned, I used to have a book of math tables that had natural logarithms but I don't recall ever using them! I doubt that anyone has ever made a table of logarithms base 5. Yes, you could use your calculator to determine the logarithms but why would you rather than just doing the multiplication on a calculator.

Well, if you insist, ln(37)= 3.6109 (that makes sense- e is about 2.7 and 3³= 27) and ln(0.785)= -.2421 (negative because 0.785 is less than one. Remember that "-1" on the common logarithm?).
The sum of those is 3.6109- .2421= 3.3688 and the "anti-logarithm" (e to that power) is 29.04 again. I still not sure what the "base" and "index" steps are unless the "base" part is taking the logarithm and the "index" part is the anti-logarithm.

As for base 5, my calculator (and I suspect yours) does not have a "logarithm base 5" key but I'm sure you know that the log base 5 of 37 is log(37)/log(5) where it does not matter which log you use:
log(37)/log(5)= 2.2436 (again, that makes sense. 5²= 25 and 37 is a bit larger than that) and log, base 5, of 0.785 is
log(0.785)/log(5)= -0.1504. That tells us that log, base 5, of 37*0.785 is 2.2436- 0.1504= 2.0931. And, of course, 37*0.785= 5^2.0931= 29.04 again.

I can't tell you how silly I feel using a calculator to determine the logarithms and do the addition when it would be much easier to use the calculator to do the multiplication itself!

 

[chemical]

oh my god thanks a lot Ivy. the help was so much appreciated!

what can i say our teacher is a bit backwards in teaching methods. we basically have to figure out log numbers without calculators and slide rules!

once again thanks