antilog on 12C

[Zac Bruce]

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 (pre-calculus) 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

[Dieter]

(08-17-2017 09:18 AM)Zac Bruce Wrote:  Evening all,

It's lunch break over here. ;-)

(08-17-2017 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.

(08-17-2017 09:18 AM)Zac Bruce Wrote:  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.

The natural antilog is e^x. The 12C has a key for that. ;-)

Dieter

[Zac Bruce]

(08-17-2017 11:33 AM)Dieter Wrote:  

(08-17-2017 09:18 AM)Zac Bruce Wrote:  Evening all,

It's lunch break over here. ;-)

The common antilog is 10^x. So simply type 10 [x<>y] [y^x].

The natural antilog is e^x. The 12C has a key for that. ;-)

Dieter

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