antilog on 12C

08172017, 09:18 AM
(This post was last modified: 08172017 09:20 AM by Zac Bruce.)
Post: #1




antilog on 12C
Evening all,
I'm hoping someone more mathematically minded than me can tell me how I can calculate the common antilog on a HP 12c, if it is indeed possible. I'm making an attempt to teach myself (precalculus) mathematics, with the aid of a self paced course at my university and Schaums Outline. I know that the common logarithm can be calculated by e.g. 'x'LN 10LN / but my feeble powers of math doesn't let me see the solution using the natural antilog in a similar way, if this is possible. Thanks in advance, Zac 

08172017, 11:33 AM
(This post was last modified: 08172017 11:36 AM by Dieter.)
Post: #2




RE: antilog on 12C
(08172017 09:18 AM)Zac Bruce Wrote: Evening all, It's lunch break over here. ;) (08172017 09:18 AM)Zac Bruce Wrote: I'm hoping someone more mathematically minded than me can tell me how I can calculate the common antilog on a HP 12c, if it is indeed possible. The common antilog is 10^{x}. So simply type 10 [x<>y] [y^{x}]. Or if you prefer a somewhat more complicated method: 10 [ln] [x] [e^{x}]. The latter may cause slight roundoff errors, e.g. x=5 returns 100 000,0005 instead of exactly 100 000. (08172017 09:18 AM)Zac Bruce Wrote: I know that the common logarithm can be calculated by e.g. 'x'LN 10LN / The natural antilog is e^{x}. The 12C has a key for that. ;) Dieter 

08172017, 10:51 PM
Post: #3




RE: antilog on 12C
(08172017 11:33 AM)Dieter Wrote:(08172017 09:18 AM)Zac Bruce Wrote: Evening all, Thanks Dieter, This was exactly what I was looking for. As I've said before; everything is easy, when you know how. It's a shame it wasn't included in the manual, the solutions handbook or in the "logarithm and exponential" training module. I was aware of the natural log/antilog. I believe we may have had a discussion involving those when I was trying to work out how to solve for 'n' in compound interest problems. I recently picked up a 35s secondhand quite cheap, and so it was actually nice to have a reason to pick that up and play around with it while my mathematically crippled brain was trying to work out what is actually going on with logarithms/anitlogs. Thanks again, Zac 

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