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

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The discussion focuses on the approximation of logarithms using the formula ½Logb(x²-1)≈logb(x), which is particularly useful for calculating logs of natural numbers, including primes. It highlights that while the approximation won't yield precise results due to the -1, it simplifies calculations by leveraging properties of logarithms. The conversation also critiques the method's efficiency, noting that looking up two logarithms to find a third may not save time. Additionally, it references Euler's manual methods for calculating logarithms, emphasizing the value of learning these techniques. Overall, the thread underscores the balance between approximation and precision in logarithmic calculations.
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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.
 
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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 ?
 
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)##
 
So instead of calculating ##\log 23## I look up ##\log 22 ## and ##\log 24## and then average ? :rolleyes:
 
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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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jim mcnamara said:
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

The link. It's dead, Jim.

-Dave K
(sorry)
 
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I had the file but no link. Thanks for the help. :smile:
 
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guifb99 said:
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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  • #11
jim mcnamara said:
I'm fixing it as we post !

The one about square roots by hand is very cool also. We are never taught this stuff.
 

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