# Revision history [back]

Well...

sage: b.solve(x).log().log_expand().solve(x)
[x == 1/2*(log(3) + 4*log(2) - log(4/5))/(log(3) - log(2))]


<Swing>"Who could ask for anything more ?"</Swing>

Maybe looking for other solutions ? (Hint, hint...)

Well...

sage: b.solve(x).log().log_expand().solve(x)
[x == 1/2*(log(3) + 4*log(2) - log(4/5))/(log(3) - log(2))]


<Swing>"Who could ask for anything more ?"</Swing>

Maybe looking for other solutions ? (Hint, hint...)

EDIT : Further hint : the logarithm is a miltivalued fincion in the complex field (i. e. not a function strictly speaking).

Well...

sage: b.solve(x).log().log_expand().solve(x)
[x == 1/2*(log(3) + 4*log(2) - log(4/5))/(log(3) - log(2))]


<Swing>"Who could ask for anything more ?"</Swing>

Maybe looking for other solutions ? (Hint, hint...)

EDIT : Further hint : the logarithm is a miltivalued fincion "multivalued function" in the complex field (i. e. not a function strictly speaking).speaking). Any time you take a log, you introduce further, possibly spurious, solutions...

Well...

sage: b.solve(x).log().log_expand().solve(x)
[x == 1/2*(log(3) + 4*log(2) - log(4/5))/(log(3) - log(2))]


<Swing>"Who could ask for anything more ?"</Swing>

Maybe looking for other solutions ? (Hint, hint...)

EDIT : Further hint : the logarithm is a "multivalued function" in the complex field (i. e. not a function strictly speaking). Any time you take a log, you introduce further, possibly spurious, solutions...

EDIT 2: Full solution, since no one seemed to see the problem :

Original problem:

,----
| b=4*3^(2*x-1)==5*4^(x+2)
----


if the members of this equations are equal, so do their logs. So we might try to solve :

,----
| Lb=b.log().expand_log()
| Lb
----

(2*x - 1)*log(3) + 2*log(2) == 2*(x + 2)*log(2) + log(5)


But the converse is not true !. More specifically :

,----
| z, z_1, z_2=var("z, z_1, z_2", domain="integer")
| (e^(x+2*I*pi*z)).maxima_methods().exponentialize()
----

e^x


Therefore, we have to consider the solutions of :

,----
| Lb2=(Lb.lhs()+2*I*pi*z_1==Lb.rhs()+2*I*pi*z_2)
| Lb2
----

2*I*pi*z_1 + (2*x - 1)*log(3) + 2*log(2) == 2*I*pi*z_2 + 2*(x + 2)*log(2) + log(5)


for any integer values of z_1 and z_2. The solutions are :

,----
| Sol=Lb2.solve(x, to_poly_solve=True)
| Sol
----

[x == 1/2*(-2*I*pi*z_1 + 2*I*pi*z_2 + log(5) + log(3) + 2*log(2))/(log(3) - log(2))]


i. e. $$\left[x = \frac{-2 i \pi z_{1} + 2 i \pi z_{2} + \log\left(5\right) + \log\left(3\right) + 2 \log\left(2\right)}{2 {\left(\log\left(3\right) - \log\left(2\right)\right)}}\right]$$

which is unique for any difference $z=z_1-z_2$.

Checking these solutions is not as direct as one could wish. But one can check that the ratio of the two members is one :

,----
| (b.rhs()/b.lhs()).subs(Sol).log().log_expand().expand().factor().exp()
----

1


One can note that the non-real roots of this equation are somehow missed by Sage (and Maxima). This is also true for giac and sympy. But Mathematica returns them:

,----
| mathematica.Reduce(b,x)
----

Element[C, Integers] && x == -((2*I)*Pi*C + 2*Log + Log + Log)/ (2*(Log - Log))
`

i. e. $$c_1\in \mathbb{Z}\land x=-\frac{2 i \pi c_1+\log (5)+\log (3)+2 \log (2)}{2 (\log (2)-\log (3))}$$

HTH,