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approximating value of decimal power of a number

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Are you asking for theory or practice?

For practical applications, it is quite easy to learn and/or calculate the log table. Start with learning that log 2 = 0.3, and of course, log 3 is approx. 0.5. Since log 4 = 2(log 2) = 0.6; log 6 = log 2 + log 3 = 0.8. So, log 5 is between log 4 and log 6, or about 0.7, and so forth.

Such methods for estimation have largely been lost with generations that have grown up with handheld calculators, but they can still be useful to know.

In your example, log (5^3.4) = 3.4 log 5 = 3.4X0.7 = 2.38
Antilog(2.38) = 100X (antilog 0.38)
Antilog (0.38) is between antilog (0.3), i.e., 2, and antilog (0.4). At this point, there are different ways to solve for antilog (0.4)

For example:
Antilog 0.4 = 0.8/2 = (log 6)/2 = 6^1/2
We know that 2.5^2 = 6.25 (i.e., 25^2 = 625)
Thus, you know that antilog 0.4 is less than 2.5, but close to it
So, an estimate for the antilog of 0.38 is between 2.0 and 2.5 and closer to 2.5

If you chose 2.4, your scratch pad calculation of 5^3.4 comes out to be 240.

My handheld calculator shows the answer as: 238

John
 
Thanks a lot, John.

Are you asking for theory or practice?

For practical applications, it is quite easy to learn and/or calculate the log table. Start with learning that log 2 = 0.3, and of course, log 3 is approx. 0.5. Since log 4 = 2(log 2) = 0.6; log 6 = log 2 + log 3 = 0.8. So, log 5 is between log 4 and log 6, or about 0.7, and so forth.

Such methods for estimation have largely been lost with generations that have grown up with handheld calculators, but they can still be useful to know.

In your example, log (5^3.4) = 3.4 log 5 = 3.4X0.7 = 2.38
Antilog(2.38) = 100X (antilog 0.38)
Antilog (0.38) is between antilog (0.3), i.e., 2, and antilog (0.4). At this point, there are different ways to solve for antilog (0.4)

For example:
Antilog 0.4 = 0.8/2 = (log 6)/2 = 6^1/2
We know that 2.5^2 = 6.25 (i.e., 25^2 = 625)
Thus, you know that antilog 0.4 is less than 2.5, but close to it
So, an estimate for the antilog of 0.38 is between 2.0 and 2.5 and closer to 2.5

If you chose 2.4, your scratch pad calculation of 5^3.4 comes out to be 240.

My handheld calculator shows the answer as: 238

John

I was just curious to know how this can be done without using a calculator. Could you please tell me how you got "6^1/2"? I'm sorry if the answer is too obvious. Thank you.

Regards
PG
 
Sorry for the late reply. I did't get notice of your post. Obviously, what I wrote is not that clear. Let me try again.

To get a good estimate of the result from your example, you need to figure out what the antilog of 0.4 is, because that is quite close to 0.38.

The log of 6 is approximately 0.8 (see calculation above).

The antilog of 0.4 can be written as antilog (0.8/2). Dividing a log by 2 is the same as taking the square root of its antilog. Thus, antilog (0.8/2) = √6 = 6^1/2

As for the square roof of 6, 25^2 = 625 (or 2.5^2 = 6.25) comes up so commonly that it is just something I memorized a long time ago. In the days before calculators, it was common to have memorized the squares up to 16 plus a few others. Failing that, we also had to learn how to extract square roots manually. I think it is nice to know it can be done, but I don't think it is worth wasting time on today.

Hope that helps.

John
 
Thank you, John.

No problem at all. There was no hurry. In the past people had really tough time doing simple calculations! But I'm sure the future generations will also think of us in the same way.

For my own reference: antilog(0.8/2)=10^(0.8/2)=sqrt(10^0.8)

Regards
PG
 
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