Re: logx y



#5

Hi Gerson,

I liked your problem because it made me cut through about 50 years of strata that overlay my school-boy days!

First I had to remind myself why ln(y)/ln(x) equals the logarithm of y to the base x.

Then it became a matter of eliminating that extra x<>y step in getting to ln(y)/ln(x).

Since (x root of y) is the same as y^(1/x), then:

((ln(x)) root of y) = y^(1/ln(x))

And taking logarithms again, ln(y^(1/ln(x))) = ln(y)/ln(x)

It was a bit of a dizzy ride, but much fun. Thanks for setting it up, and for the interesting link.

Cheers, Tom


#6

The message above was in response to Gerson's thread below!

#7

Another interesting thing:

Shortly after I came out with this, Joe Horn became somehow obsessed with the accuracy of the method. He carried out a complete study on the subject:

http://groups-beta.google.com/group/comp.sys.hp48/browse_frm/thread/85091c9dfcd2465/2b3bb175a40d0482?q=LN+XROOT+LN&rnum=2&hl=en#2b3bb175a40d0482

I think it is just something to be used if you're in desperate need of memory.

Cheers, Gerson


#8

Gerson,

Congratulations to you on your really neat discovery, and thanks for sharing it with us here. I find it fascinating how the most elementary ideas of mathematics can so often be viewed from fresh perspectives, with surprising results. And sometimes the calculator is an aid in doing this.

Cheers, Tom


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