Solving Exponential Equations  Printable Version + HP Forums (https://archived.hpcalc.org/museumforum) + Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum1.html) + Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum2.html) + Thread: Solving Exponential Equations (/thread148181.html) 
Solving Exponential Equations  Hal Bitton in Boise  03112009 Hi folks,
When I solved this equation on my TI89 emulator, and noted that only the principle solution of 2 was returned, with no mention of the complex solutions. I was just wondering if there is a reason for this seemingly incomplete solution from the TI (perhaps the complex series is considered superfluous?), and is there a way to get it to deliver the complex solutions? Re: Solving Exponential Equations  MacDonald Phillips  03112009 The TI89 and TINSpire CAS use cSolve() to solve for complex (and real) roots.
Re: Solving Exponential Equations  Chuck  03112009 What's odd is that the 50G doesn't make it a little simpler: x = (n 2 pi i)/ln(3) + 2 (for any integer n). Edited: 11 Mar 2009, 7:40 p.m.
Re: Solving Exponential Equations  Michael de Estrada  03112009 I think you're asking too much of it. It would take a pretty sophisticated simplification algorithm to figure out that:
e^(i pi/2) = i
Re: Solving Exponential Equations  Marcus von Cube, Germany  03122009 Here is what the Nspire CAS Mac/PC Software returns:
The second line was created by just selecting >Rect from the menu, Ans has been replaced by the calculator in the history view. BTW, the HP50g returns essentially the same solution:
I used the SOLVEVX softkey (F5) after a DUP to have both expressions on the screen. I just couldn't find a simple way to split the solution.
I couldn't get it to work on my HP40gs: SOLVEVX(3^{X}=9) returns an error: Non unary operator. Edited: 12 Mar 2009, 6:48 a.m.
Re: Solving Exponential Equations  Steve Keeley  03122009 Chuck wrote:
Quote:
If you delete the "X=" leaving
2+(2*i*n1*pi)/LN(3)
Re: Solving Exponential Equations  Karl Schneider  03132009 Quote:
Hmm, this doesn't work on my HP49G  I'm still seeing 2*n1*pi [ pi] [sarcasm]Maybe flag 279 isn't set set correctly...[/sarcasm] Seriously, the question everyone ought to answer for themselves is... "Can I solve this on pen and paper?"  KS
Re: Solving Exponential Equations  C.Ret  03132009 May be du to the 'general solution' versus 'principal value' mode setting. 34 SF '3^X=9' X ISOL returns 2.which is the ‘principal value’ or obvious value. 34 CF '3^X=9' X ISOL returns (2.1972+2*pi*n1)/1.0986which is the ‘general solution’ solution. The ‘principal value domain’ is definite for each analytical function of your calculator and is give on any good user manual by a graph. I am not sure that the ‘principal value domain’ definitions are exactly the same on any calculators, variations may exist between HP and TI, or between models from the same brand (due to technology evolution and mathematical choices). As it can be seen there, the HP28S/C don’t give a pure analytical solution with ISOL command such as HP49 or HP50. To get it the same form as HP50g, a HP28C/S user have to follow the following method which is close to the ‘by hand’ resolution. Principal(real) solution To avoid resolution of an unknown x at power of a number a, one may use the logarithm to transform the equation from a^{x} = b to ln(a^{x}) = ln(b) assuming a^{x} is strictly positive (principal or obvious solution) and NOT complex (general solution). Remember the logarithm property of transforming products into additions: ln(a^{x}) is transformed into x.ln(a). Thus, the equation is now writen as: x.ln(a) = ln(b). As far as a is not unity, ln(a) is not zero, so we get: x = ln(b)/ln(a). N.A.: ‘3^X=9’ give principal solution ‘X=2’ since ln 9 = 2.ln 3. We also have to verify that 3^{2} > 0 and 3=/=1, which is true, so ‘X=2’ is a valid ‘principal’ solution. The general (complex) solution As the principal one, the general solution is obtained by using logarithm, but considering X as a complex. If x is complex, we know that it can be expressed as a function of its argument tand module (absolute value) r: x = r.e^{i.t} The equation is now : a^{r.ei.t} = b to which may be transform into by using logarithm : [italic]ln(a^{r.ei.t}) = ln(b) r.e^{i.t}. ln(a) = ln(b) r.e^{i.t} = ln(b)/ln(a) To found the general solution of 3^{x}=9, we have to found all the possible value of (r,t) which satisfy the equation: r.e^{i.t} = 2 Two complex are equal when their modules (absolute value) and arguments are both simultaneously equal. Since 2 = 2+0.i = 2.e^{0i}, we have to solve the system:
By definition, e^{i.t} is cos(t)+i.sin(t):
Equation 2 is true for any t = ... 6pi, 4pi, 2pi, 0, 2pi, 4pi, 6pi, ... We can express that t have to be zero modulo 2pi or t = 2.pi.n Thus x = [ ln(b) + 2pi.n.i ]/ln(a) Conclusion, ‘general solutions’ are all complex number of module (absolute value) 2 and argument 0 modulo (2pi). This can be expresse as 3^{x}=9 for any x = 2 + 2pi*i*n/ln(3) where n is any integer. The ‘principal value’ solution is found at n = 0 where x =2
Re: Solving Exponential Equations  Karl Schneider  03142009 "C.Ret"  Thank you for showing about flag 34 on the HP28C/S, and its effect on ISOL. Here is an old thread that suggests increasingly better performance of ISOL on the HP28, HP48, and HP49 for a fairly straightforward problem: http://www.hpmuseum.org/cgisys/cgiwrap/hpmuseum/archv014.cgi?read=70012 Thanks also for writing your approach for obtaining the solution manually. The solution seems fairly obvious in retrospect, after it has been methodically derived by this or another method.  KS
Edited: 16 Mar 2009, 12:09 a.m.
