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How to obtain algebraic solution of the logarithmic equation

expression=-b n + Exp[t] lambda n
Solve[ expression == 0, t]

The output should be $$t=Ln \left(\frac{b}{lambda}\right)$$, but I get.

{{t -> ConditionalExpression[2 I \[Pi] C[1] + Log[b/lambda], 
    C[1] \[Element] Integers]}}
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1 Answer 1

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Tell Mathematica (Element[t,Reals]) what you are looking for (and change e^t to Exp[t] )

expression = -b  + Exp[t] lambda  
Solve[expression == 0, t, Reals]
{{t -> ConditionalExpression[Log[b/lambda], (b > 0 && lambda > 0) || (b < 0 && lambda < 0)]}}
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  • $\begingroup$ what I posted is an MWE, it is not the real problem. In the real problem, I do not know the final answer. So, I can not tell Mathematica what I am looking for. My question would be then "If I do not know the solution of an equation, how to find this solution using Mathematica?" $\endgroup$
    – JuanMuñoz
    Commented Dec 11, 2020 at 14:08
  • $\begingroup$ What I meant is if you know you're looking for a real solution use Solve[…, Reals] $\endgroup$ Commented Dec 11, 2020 at 14:41
  • $\begingroup$ This response is simple and goes directly to the question that was asked. If it is not adequate, that indicates something amiss in the question. $\endgroup$ Commented Dec 11, 2020 at 14:46
  • $\begingroup$ @UlrichNeumann Right thanks you!!!! I wonder, how can I extract the solution and save it as a new variable x_new=Solve[ expression == 0, t] $\endgroup$
    – JuanMuñoz
    Commented Dec 11, 2020 at 15:09
  • $\begingroup$ @JuanMuñoz Perhaps ` tnew=(t /. sol)[[1, 1]]` as a workaround? $\endgroup$ Commented Dec 11, 2020 at 15:59

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