B Log(x), an easy and useful way to calculate it

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½Logb(x2-1)≈logb(x)

This is an easy and useful way to calculate the log of any natural number, including primes, it won't ever give a precise result, obviously (because of the -1), but as "x2-1" will always have divisors smaller than "x", you can easily calculate the approximation by using the property that logc(a*b)=logc(a) + logc(b).

It could actually just be n-1logb(xn-k)≈logb(x) but it would be harder to calculate depending on which "k" or which "n" you use and won't be as useful for school purposes.

Obviously, since ½logb(x2)=logb(x), the closer "k" is to 0, the more precise the result will be. And the closer "n" is to infinity, also the more precise the result will be, since the value of "k" will become less and less significant as "xn" gets bigger.

*It's good to notice that for a "k" bigger than one, it will fail miserably.
 

BvU

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Hello guif,

How would you actually go about to calculate such a logarithm ? Can you give an example ? What if e.g. x = 23 ?
 

Stephen Tashi

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The general idea is the approximation ##f(x) \approx \frac{f(x-1) + f(x+1)}{2} ##.

for ##x > 1## , ##\log_b(x^2 -1) = \log_b((x-1)(x+1)) = \log_b(x-1) + \log_b(x+1)##
 

BvU

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So instead of calculating ##\log 23## I look up ##\log 22 ## and ##\log 24## and then average ? :rolleyes:
 
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jim mcnamara

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You still have to look up logarithms with your method. Basically if you have to look up two logarithms to get a third, you could have saved yourself work by simply just looking up the log you wanted first off. You are not saving anything, really.

Euler developed a simple way by hand using 4 properties of properties of logarithms. So learn how to do logs by hand, I did, it's fun.

Try a simple google search.
 
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You still have to look up logarithms with your method. Basically if you have to look up two logarithms to get a third, you could have saved yourself work by simply just looking up the log you wanted first off. You are not saving anything, really.

Euler developed a simple way by hand using 4 properties of properties of logarithms. So learn how to do logs by hand, I did, it's fun.:
https://www.fiziko.bureau42.com/teaching_tidbits/manual_logarithms.pdf [Broken]
The link. It's dead, Jim.

-Dave K
(sorry)
 
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jim mcnamara

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I had the file but no link. Thanks for the help. :smile:
 
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½Logb(x2-1)≈logb(x)
For large x, ##x^2 - 1 \approx x^2##, so ##\log_b(x^2 - 1) \approx \log_b(x^2) = 2 \log_b(x)##. Your equation above comes immediately from this one.

For example, with x = 25, ##\frac 1 2 \log(25^2 - 1) \approx 1.397592## and ##\log(25) \approx 1.39794##.
 
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Here I found way to find Solution Without Calculator. I should say to the approximate logarithm(base 10) of any number, you need to remember the above three values. And this method I used and worked for me. I am sharing it here.
 

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