# Revision history [back]

[ This smells homework. Therefore, some hints... ]

Ahem ! Your equation has, indeed one real solution... and 18 complex solutions. Its structure will become more transparent if you accept to replace its float coefficient by exact numbers and solve it symbolically...

The numerical values should also give you some hints :

sage: [(E.subs(s).rhs()-E.subs(s).lhs()).n().abs() for s in solve(0.1*1==1*e^(-0.38*t),t)]
[8.88516071398867e-16,
8.97447312850170e-16,
7.34788079488409e-17,
8.93565536393413e-16,
8.96581565195064e-16,
9.00254100846723e-16,
9.04575147216186e-16,
9.09535461462144e-16,
9.15124648438264e-16,
2.22176931601818e-16,
1.97899526057791e-16,
5.39806219580408e-16,
1.49555718205910e-16,
1.25570576985110e-16,
1.01827474616583e-16,
7.85462105727423e-17,
5.63026431885206e-17,
3.70172377524261e-17,
2.77555756156289e-17]


[ This smells homework. Therefore, some hints... ]

Ahem ! Your equation has, indeed one real solution... and 18 complex solutions. Its structure will become more transparent if you accept to replace its float coefficient by exact numbers and solve it symbolically...

The numerical values should also give you some hints :

sage: [(E.subs(s).rhs()-E.subs(s).lhs()).n().abs() for s in solve(0.1*1==1*e^(-0.38*t),t)]
[8.88516071398867e-16,
8.97447312850170e-16,
7.34788079488409e-17,
8.93565536393413e-16,
8.96581565195064e-16,
9.00254100846723e-16,
9.04575147216186e-16,
9.09535461462144e-16,
9.15124648438264e-16,
2.22176931601818e-16,
1.97899526057791e-16,
5.39806219580408e-16,
1.49555718205910e-16,
1.25570576985110e-16,
1.01827474616583e-16,
7.85462105727423e-17,
5.63026431885206e-17,
3.70172377524261e-17,
2.77555756156289e-17]


EDIT to answer your further question : another way to solve this is, as I told already, to use an exact ring, i. e. replace 0.1 by 1/10 and 0.38 by 38/100 (or 19/50, according to your tastes). So your problem becomes :

sage: var("t")
t
sage: E=1/10-1*e^(-38/100*t) ## Left-hand - right-hand
sage: S=solve(1/10-1*e^(-38/100*t),t) ## (Exact) solutions
## Check by substitution that all the proposed solutions are indeed solutions.
sage: [E.subs(s).trig_expand().expand() for s in S]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]


And your real solution is :

sage: [s for s in S if s.rhs().is_real()]
[t == 50/19*log(10)]


Clearer ?

[ This smells homework. Therefore, some hints... ]

Ahem ! Your equation has, indeed one real solution... and 18 complex solutions. Its structure will become more transparent if you accept to replace its float coefficient by exact numbers and solve it symbolically...

The numerical values should also give you some hints :

sage: [(E.subs(s).rhs()-E.subs(s).lhs()).n().abs() for s in solve(0.1*1==1*e^(-0.38*t),t)]
[8.88516071398867e-16,
8.97447312850170e-16,
7.34788079488409e-17,
8.93565536393413e-16,
8.96581565195064e-16,
9.00254100846723e-16,
9.04575147216186e-16,
9.09535461462144e-16,
9.15124648438264e-16,
2.22176931601818e-16,
1.97899526057791e-16,
5.39806219580408e-16,
1.49555718205910e-16,
1.25570576985110e-16,
1.01827474616583e-16,
7.85462105727423e-17,
5.63026431885206e-17,
3.70172377524261e-17,
2.77555756156289e-17]


EDIT to answer your further question : another way to solve this is, as I told already, to use an exact ring, i. e. replace 0.1 by 1/10 and 0.38 by 38/100 (or 19/50, according to your tastes). So your tastes), so you will work with rationals (which have an exact representation in Sage) in place of "floats", which are limited-precision approximations).

sage: var("t")
t
sage: E=1/10-1*e^(-38/100*t) ## Left-hand - right-hand
sage: S=solve(1/10-1*e^(-38/100*t),t) ## (Exact) solutions
## Check by substitution that all the proposed solutions are indeed solutions.
sage: [E.subs(s).trig_expand().expand() for s in S]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]


And your real solution is :

sage: [s for s in S if s.rhs().is_real()]
[t == 50/19*log(10)]


Clearer ?