tag:blogger.com,1999:blog-2433841880619171855.post3303724325444395841..comments2024-03-27T21:09:44.320+00:00Comments on Pat'sBlog: On This Day in Math - February 1Unknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-2433841880619171855.post-69092407783048005022014-02-01T13:52:33.993+00:002014-02-01T13:52:33.993+00:00I think this 1/e is also not correct. The best ap...I think this 1/e is also not correct. The best approximation between them I know of is <br /> <br /> Napier's Log(x) appx = 9999999.5 (7 ln(10) - ln(x) )<br />Pat's Bloghttps://www.blogger.com/profile/15234744401613958081noreply@blogger.comtag:blogger.com,1999:blog-2433841880619171855.post-54146106082068685322014-02-01T08:37:28.508+00:002014-02-01T08:37:28.508+00:00If I remember correctly, Napier's logarithms w...If I remember correctly, Napier's logarithms were in a base roughly equal to 1/e.<br /><br />Since the base is negative, they went "the wrong way"....mau.https://www.blogger.com/profile/09641196427325175260noreply@blogger.com