Re: Solving Exponential Equations Message #8 Posted by C.Ret on 13 Mar 2009, 8:00 a.m., in response to message #7 by Karl Schneider
May be du to the 'general solution' versus 'principal value' mode setting.
On the HP28C/S, this is drive by the flag 34: 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)/1.0986 which 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: r = 2
 e^{i.t} = 1
By definition, e^{i.t} is cos(t)+i.sin(t): r = 2
 cos t + i.sin t = 1
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)
x = [ 2.ln(3) + 2pi.n.i ]/ln(3)
x = 2 +2pi.n.i/ln(3)
x = (2, 5.7192n)
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
