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#1 2006-10-14 18:57:20

ganesh
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Registered: 2005-06-28
Posts: 13,428

Linear Interpolation

Logarithms are a very useful tool in calculations.

By remembering the log to base 10 values of just three numbers, the approximate log value of any number can be found.

All you have to remember is (all approximate values)
log(10)2=0.3010
log(10)3=0.4771
log(10)7=0.8451

log(10)4 would be 2log(10)2
log(10)5 would be 1 - log(10)2
log(10)6 would be log(10)2+log(10)3
log(10)8 would be 3log(10)2
log(10)10 would be 2log(10)3
log(10)10 is 1.

Illustration : To find the log(10)23,
we know log 20 is 1.3010,
log 25 is 2log(10)5, log(10)5 is 0.6990
Therefore, log(10)25 would be 1.3980
23 is 3/5 between 20 and 25.
Find the difference between log(10)25 and log(10)20. It is 0.0970
3/5 is 6/10, therefore, 6/10 of 0.0970 is 6*0.0097
that is 0.0582.
Add this to log(10)20.
You get 1.3592. This would be approximately log(10)23.
The scientific calculator gives the value as 1.3617.
Although there's a 0.0025, the percentage difference is only 0.18%.
This works much better for higher numbers.
Try finding the log(10)value of 82, you can calculate the log(10) values of 81 and 84 with what you remember. log(10)82 would be 1/4 of the difference added to log(10)81.
I found this method helpful.


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#2 2007-01-24 17:11:11

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Linear Interpolation

Hey uh, common log 82 is 1/2.9639 of the way between log81 and log84, which is close to 1/3.
Probably just a typo.


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#3 2007-02-03 21:46:06

ganesh
Moderator
Registered: 2005-06-28
Posts: 13,428

Re: Linear Interpolation

Thanks for correcting me, John,
yes, that was a typographic error.


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