Hidden figures: 2.303, slide rules and classrooms mired in the last century

A five -place table of logarithms from my dad's CRC Handbook of  Mathematics (why is that set of values circled?) and a circa  1958 Hemmi 257 slide rule designed for chemical calculations.

 Wonder why random values of 2.303 are "hidden" in formulae? To make them easier to use with a slide rule.¹ 

A slide rule?  The last slide rule slid out the door of Keuffel & Esser in 1975 (they sent their engraving equipment to the Smithsonian).  You can still find them, used and even new - still packaged up to sell to engineers and scientists.  The Oughtred Society has a online museum, as well.

We still have my mother-in-law's K&E, in it's leather case with her name impressed into it.  Family history says she bought it with the money she earned tutoring Jackie Robinson in chemistry at UCLA.

I have an essay out in this month's Nature Chemistry, "It figures", about how the computational tools we use shapes what we teach and not necessarily in good ways. Given that slide rules were obsolete by the time many of my student's parents were born, why does their use still linger in general chemistry book?  (The 2.303's in texts are lowly going away. I checked texts running back about a decade.)

More critically to my mind why, several decades after  digital computing tools became ubiquitous on college campuses do many physical chemistry texts eschew any discussion of numerical techniques for solving the rate equations for a chemical reaction?  I suspect the chasm between the computational tools used in the field and those used in the classroom is a result of apathy. We teach what we learned as we learned it.  As I note in the article, I don't think it is defensible on intellectual grounds.

Don't know how to use a slide rule?  It's fun, it's geeky. No need to buy one to play, check out this simulator and the instructions at Nature Chemistry!

You can read the article here:  http://rdcu.be/sY5Q

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1.  2.303 is the natural log of 10. To change the base of logs recognize that
x = b^log_bx
so
ln(x) = ln(10^log₁₀x)
ln(x) = log₁₀x ln(10)
ln(x) =(log₁₀x)(2.303)
ln(x) = 2.303(log₁₀x)