Evaluation of anti-log

[EdClymer]

Greetings from England

In my kind of work, log function is very unusual.

I have found that log(number)= works fine - in that answer is correct.

y: 4.601 log = 0.66285

But now I need to evaluate the anti-log...can be easily done on pocket
calculator (via inverse/shift); but how is it done in Mathcad?

By experimentation...

10^log = 4.601 works, but is it correct procedure?; any neater solution?

I use Mathcad 2000.

Regards

Ed





Ed Clymer
Resinfab & Associates
England

 

[IRstuff]

That is correct, since that is, by definition, what a logarithm is, i.e., log(x) is the value by which 10^log(x) = x

TTFN
faq731-376

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Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[Occupant]

Or create your own function: alog(x):10^x where "x" would your previous log()

 

[EdClymer]

Thanks to IR stuff and Occupant.

IR stuff is quite correct, back to basics.

Occupant has neat solution, I will give that a try.

Help appreciated.

Regards

Ed

Ed Clymer
Resinfab & Associates
England