How do you express the decimal numbers like 10,000 & 1,000,000 as a power of 10 ???

 

10 000 = 10^4

1 000 000 = 10^6

Count the zeroes, look at the power; notice the pattern?

 

You can express any number at all as a power of 10 using logarithms. In the case of powers of ten, the logarithm happens to be an integer. It is also the case that exponetiation and taking the logarithm are inverse functions of each other.

 

To find what power of 10 equals a specific number X (any real number), you have \(10^Y = X\). Take the base 10 log of both sides of the equation: \(\log{10^Y} = \log{X}\). Then using \(\log{(10^A)} = A\), we have: \(Y = \log{X}\).

So, for X= 5.296, for instance, we have \(Y = \log{(5.296)} = 0.72394797643163\)

Then, you'll find that \(10^{0.72394797643163} = 5.296\)

 

Apologies for my simple reply! I misinterpreted the question.

 

Apologies for my simple reply! I misinterpreted the question.

I don't think so -- your reply was just fine for the OPs question. We were just trying to point out the much richer structure of logarithms and exponents as an additional FYI for the curious among us.

There is also the possible ulterior motive of demonstrating our extended body of knowledge and experience. Do 'ya think?

 

Papabravo,

You can express any number at all as a power of 10 using logarithms.

But not negative numbers, unless you want to tangle with complex numbers.

zgozvrm,

To find what power of 10 equals a specific number X (any real number), you have . Take the base 10 log of both sides of the equation: .

You can be more general than that. Suppose I want the binary logarithm of 1024. Do ln(1024)/ln(2) = 6.9314718056/0.69314718056 = 10 . Or use a different log base for calculation; log(1024)/log(2) = 3.0102995664/0.301029995664 = 10 . But don't try to find the logarithm of a negative number unless you can accept complex numbers.

Ratch

 

You can be more general than that. Suppose I want the binary logarithm of 1024. Do ln(1024)/ln(2) = 6.9314718056/0.69314718056 = 10 . Or use a different log base for calculation; log(1024)/log(2) = 3.0102995664/0.301029995664 = 10 .

Of course, you can use any logarithm (natural, base 10, base 7, etc.), but since we're already trying to solve for \(10^x\), it just makes sense to use the base 10 logarithm; it simplifies the calculation slightly.

 

For negative numbers you just put a minus sign in front of both sides as in

 y =  10^x
-y = -10^x

Where is the need for compex numbers?

They do come about if you want to say that there is a Laurent series that conveges to ln(z), that is analytic in some neighborhood, except for a removeable singularity at the origin. There is also the matter of restricting arg(z) so tha the function does not become multivalued.

 

Papabravo,

For negative numbers you just put a minus sign in front of both sides as in

Code:
y = 10^x
-y = -10^xWhere is the need for compex numbers?

What is log10(-3)? How are you going to put a minus sign in front of that?

Answer is the complex number (0.47712125472,1.36437635384). In other words, 10^(0.47712125472,1.36437635384) = -3 .

Ratch

 

We are talking about two different problems. The original problem was to represent a real number as a power of 10.

The second problem is to take the logarithm of a negative number. It is a related problem and your exposition is correct. I have no argument with anything that you've said. In the case of trying to take the inverse of the equation -y = -10^x, order of operations is crucial, and the operations are NOT commutative. This state of affairs happens all the time.

 

Papabrave,

...The original problem was to represent a real number as a power of 10.

Yes, but negative numbers are real numbers, too. I was just being pedantic in pointing out that logs of negative numbers are not real like positive numbers are.

Ratch

 

I read your articles here with interest and so I quickly set about writing you a very tiny peace of software to make life easier using the Java specification

If you can run java then here is a small program attached

Have fun with this for those of you who would like to see how we do this here I've also attached the code

I'm using a single frame launched from main.java

A couple of textfields which pass a double value to the variables then the Math class to get my answers when buttons receive events I've also included the JOption class so that in the event of an error been thrown the textfields are reset to "" The process can be repeated any number of times

To run the application simply open a dos box after downloading the attachments to a folder of your choice CD or Change Directory to that folder using the dos commands and then type

java -jar Logarithms.jar Up should come a window and away you go

Enjoy

 

Thanks Harrington. That makes life a little 'quicker'.

 

Before calculators, there was a standard reference book - 'Logarithmic and Other Tables' - that's one way of finding the log of any number?

(Or, even less accurately, a Slide Rule)

Still got both...

 

For folks who went to college before calculators were invented, most of you will no doubt remember what a pain it was to have to do e.g. calculations to 5 or more significant figures using log tables -- or, even worse, by long-hand techniques.

For a humbling experience, set yourself up a reasonably complicated problem including multiplications, divisions, roots/powers, and some trig functions. Then figure out the answer using e.g. an old CRC math tables book. Oh, and time yourself too.

No fair using a calculator or computer, either. You have to get the answer with those old methods and be confident of the result.

Every decade or so I'll try such a task and marvel at how the old skills have (nearly) completely atrophied. I suppose we could relearn the task if necessary, but it's horrible drudgery that no one would want to go back to.

I wonder if anyone younger than their 50's learned such things -- did they continue to teach such skills after calculators came out?