### Logarithms

The other day, when I was wanting to code some financial formulae in F#, I got side-tracked and derived something that's not intuitive to me:

Can that really be? I didn't see an obvious mention of this in the Logarithms article on Wikipedia, nor in the pre-calc book we have at home. So maybe I screwed up the math? I tried some test cases with F#, by defining a function that, given a

> let logcheck b x = (Math.Log(x, b), (1.0 / Math.Log(b, x)));;

val logcheck : float -> float -> float * float

> logcheck 2.0 3.0;;

val it : float * float = (1.584962501, 1.584962501)

> logcheck 5.0 11.2;;

val it : float * float = (1.501091629, 1.501091629)

> logcheck 7.382 123.123;;

val it : float * float = (2.407742101, 2.407742101)

The equivalence seems to be checking out, so maybe I did the math right. As it turns out, this equivalence

Book ordered:

Can that really be? I didn't see an obvious mention of this in the Logarithms article on Wikipedia, nor in the pre-calc book we have at home. So maybe I screwed up the math? I tried some test cases with F#, by defining a function that, given a

*b*and an*x*, evaluates both sides of that last equation, resulting in a tuple of two floats:> let logcheck b x = (Math.Log(x, b), (1.0 / Math.Log(b, x)));;

val logcheck : float -> float -> float * float

> logcheck 2.0 3.0;;

val it : float * float = (1.584962501, 1.584962501)

> logcheck 5.0 11.2;;

val it : float * float = (1.501091629, 1.501091629)

> logcheck 7.382 123.123;;

val it : float * float = (2.407742101, 2.407742101)

The equivalence seems to be checking out, so maybe I did the math right. As it turns out, this equivalence

*is*noted in the middle of a Wikipedia page. But what the heck does it mean?Book ordered:

## 1 Comments:

At 12:55 AM , Anonymous said...

One way to understand basic logs is: A function taking two inputs and returning the required exponent to take the first number to the second number.

Example: Log(2, 8) = 3 since 2^y = 8 requires that y = 3.

When viewed this way, it is obvious that Log(8, 2) = 1/3 = 1/Log(2, 8).

Regards,

CarlMon

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