Sorry, my post was in "pseudo-mathematics," a dialect of math you can only use when you have nothing but a computer keyboard to work with. 8)

The keystrokes are close to what Chris has in his program:

n

g LOG

2

g LOG

/

Here "n" is a number you key in and "g" is the blue shift key. So g LOG is g followed by the third key from the left in the upper row, marked "10^X". For example, to compute the logarithm base 2 of 16, you would do "16 g LOG 2 g LOG /" and see the answer of 4.000 (in FIX 4) in the display. Chris suggests writing a program if you do this calculation frequently. You would first do "g R/S" to enter program mode, then type Chris's keystrokes, then type "g R/S" again to leave program mode. Once you did that, pressing the 10^X key with user mode set ("f RCL", where f is the gold shift key, will toggle user mode) will run your program. The keystrokes to compute the logarithm base 2 of 16 would then be "16 10^X"

If this all makes sense, and the following *doesn't* make sense, then just ignore the stuff you don't understand.

The program and keystroke solution above would work just as well using the *second* key from the left in the top row, instead of the third. That's because, mathematically, the logarithm to any base of a number, divided by the logarithm to the same base of another number, will produce the logarithm to the second number's base of the first number. English has it's limitations in describing mathematical ideas, so that probably sounds confusing. Symbolically:

log_{B1}**n**

---------- = log_{B2}**n**

log_{B1}**B**_{2}

Good thing I found the subscript function. It helps emphasize that this is a math formula and not a 15C program.

So in the first case, the third key from the left (blue shifted) is the logarithm base 10, so B_{1} is ten. But you could use the (blue shifted) *second* key from the left just as well, because that's the logarithm base *e*, where *e* is a magic number that is useful in many areas of math. So B_{1} in that case is *e*. But everything works because you use the same base in both logarithms, and that matches the formula.

Hope that helped.

*Edited: 9 Aug 2005, 3:15 p.m. *