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

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1. Dec 20, 2016

guifb99

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.

2. Dec 20, 2016

BvU

Hello guif,

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

3. Dec 20, 2016

Stephen Tashi

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

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

4. Dec 20, 2016

BvU

So instead of calculating $\log 23$ I look up $\log 22$ and $\log 24$ and then average ?

5. Dec 20, 2016

Staff: Mentor

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.

6. Dec 20, 2016

dkotschessaa

-Dave K
(sorry)

Last edited by a moderator: May 8, 2017
7. Dec 20, 2016

dkotschessaa

8. Dec 20, 2016

Staff: Mentor

I'm fixing it as we post !

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9. Dec 20, 2016

Staff: Mentor

10. Dec 20, 2016

Staff: Mentor

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 immediated from this one.

For example, with x = 25, $\frac 1 2 \log(25^2 - 1) \approx 1.397592$ and $\log(25) \approx 1.39794$.

11. Dec 20, 2016

dkotschessaa

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