Prev: a crafty solution to a key problem
Next: UBASE: The Bug ... still there after all these years!
From: Anssi Saari on 15 Jul 2010 15:29

I was recently faced with a little problem, to solve log(x, 64) = 10
for x (i.e. base x logarithm of 64 equals 10).

All I got from my old HP48SX was the approximate value and at the time
I got stuck with e^(ln(64)/10) and didn't get to the simplest solution
(x=2^(3/5) or XROOT(5,8)).

Would a HP50G have served me better here? I tried a 50G in Debug4x,
but I had to admit defeat, it seems to be quite different from my old
48SX...

From: John H Meyers on 15 Jul 2010 18:23
On 7/15/2010 2:29 PM, Anssi Saari wrote:

> I was recently faced with a little problem, to solve log(x, 64) = 10
> for x (i.e. base x logarithm of 64 equals 10).
>
> All I got from my old HP48SX was the approximate value and at the time
> I got stuck with e^(ln(64)/10) and didn't get to the simplest solution
> (x=2^(3/5) or XROOT(5,8)).
>
> Would a HP50G have served me better here? I tried a 50G in Debug4x,
> but I had to admit defeat, it seems to be quite different from my old
> 48SX...

To the HP48SX, add some extra commands and "forms" from the HP48GX,
add "MetaKernel" (e.g. the Filer) and more symbolic algebra (CAS),
and you have the HP49/50 series.

CASCFG

'LN(64)/LN(X)=10' 'X' SOLVE EVAL ==> 'X=EXP(3*LN(2)/5)'

This is equivalent to the other expressions you wrote,
or to 'X=8^(1/5)'

By the way, what's the precise definition of "simplest,"
and is there only one possible correct answer,
or are all equivalent answers still correct?


[r->] [OFF]
From: Wes on 16 Jul 2010 02:54
On Jul 15, 10:29 pm, Anssi Saari <a...(a)sci.fi> wrote:
> I was recently faced with a little problem, to solve log(x, 64) = 10
> for x (i.e. base x logarithm of 64 equals 10).
>
> All I got from my old HP48SX was the approximate value and at the time
> I got stuck with e^(ln(64)/10) and didn't get to the simplest solution
> (x=2^(3/5) or XROOT(5,8)).
>
> Would a HP50G have served me better here? I tried a 50G in Debug4x,
> but I had to admit defeat, it seems to be quite different from my old
> 48SX...

I tell my students that the best way to solve this kind of a problem
is to set their calculators aside and just do the algebra. :-)

log_x (64) = 10
x^10 = 64 (def. of log)
x = 64^(1/10) (positive only, unless you're okay with negative
bases)
x = (2^6)^(1/10)
x = 2^(6/10)
x = 2^(3/5)

But if the point was to do it on the calculator, since the HP's don't
have a log_base function, you'll have to rewrite it. You could
rewrite it as

ln(64)/ln(x) = 10

but if you're going to have to rewrite it anyway, you might as well
write it as

x^10 = 64

-wes
From: Anssi Saari on 17 Jul 2010 06:55
Wes <wjltemp-gg(a)yahoo.com> writes:

> I tell my students that the best way to solve this kind of a problem
> is to set their calculators aside and just do the algebra. :-)

Sure. But this was a time limited entrance examlet. 10 problems in 3
hours and also needed to write a short 200 word essay in English to
prove I can. Since it's been about 15 years since my last math class,
it really would've helped to have a little more help from the calc.
Got in anyways, so I'll be going to night classes about mobile
programming this fall.

Anyways, I figured post fact it's also easy to convert directly to
base 2 logarithm, since 64 is a power of 2 and that gets one to log(2,
x) = 3/5 in about two steps.
 | 
Pages: 1
Prev: a crafty solution to a key problem
Next: UBASE: The Bug ... still there after all these years!