Converting Logs

[Numbers]

I was actually rather interested in the math content of an earlier thread, so if no one minds I would like to re-visit that topic.

The discussion was about converting logs from one base to another, and especially to base e. I wasn’t able to follow the example given, but have come up with the following:

If y = a^x, then log(y) = xlog(a)

So log(y) / log(a) = x

So to convert from log_10 to log_e I get

Log_10(y) / Log_10(e) = base e log of y

Is this correct?

[Mystwalker]

If I haven't misread something, that is totally correct (and the way to use when a calculator only has log_10).

[Numbers]

Thank you very much.
Maybe I was able to follow the other example after all.

[dsouza123]

And e (in base 10) is 2.718281828 to 9 places after the decimal point.
To get e take the inverse natural log of 1,
using the Windows XP calculator in Scientific mode press 1 Inv ln

[Numbers]

dsouza123,
That would account for my error. I had been using the series

1 + 1/1! + 1/2! + ... 1/12!

which gives 2.7182818283, and this resulted in my experiments with my conversion formula to sometimes give results that did not quite agree (although only in the ninth or tenth decimal place). Inv ln(1) is much more accurate (not to say simpler) and consistently gives the correct result, thank you very much.

[Glenn Leider]

I like "1 Inv ln" for e too. As for a quicker converging expression than

1 + 1/1! + 1/2! + 1/3! + 1/4! + ... + 1/12! try

1 + 2/1/1 - 2/1/7 + 2/7/71 - 2/71/1001 + 2/1001/18089 - 2/18089/398959 +

The pattern: 1 + 6x1 = 7, 1 + 10x7 = 71, 7 + 14x71 = 1001,
71 + 18x1001 = 18089, 1001 + 22x18089 = 398959, etc.

It's based on this continued fraction:

e = 1 + 2/(1 + 1/(6 + 1/(10 + 1/(14 + 1/(18 + 1/(22 + ...))))))