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[?] Use Solve to get an integer solution?

Ray Lawicki, COLLEGE
Posted 10 years ago

How do I get Mathematica to solve 3^(2x-1)=27 . if i type 

Solve[3^(2x-1)==27,x]

I get a complex response which is obviously wrong since the answer is 3?
POSTED BY: Ray Lawicki
8 Replies
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Try:

 FindInstance[3^(2 x - 1) == 27, x, Integers]
 (*{{x -> 2}}*)
POSTED BY: Mariusz Iwaniuk
Ray Lawicki
Ray Lawicki, COLLEGE
Posted 10 years ago

thanks for the hint. it seems to work with solve, reduce and findinstance as long as the expression is followed by "integers" or "reals" which i neglected to add to the problem. Solve[3^(2 x - 1) == 27, x, Integers] {{x -> 2}}
POSTED BY: Ray Lawicki
Ray Lawicki
Ray Lawicki, COLLEGE
Posted 10 years ago

thanks for the solution i did not realise that restricting the domain to real was the trick
POSTED BY: Ray Lawicki

There are actually an infinite number of answers, but only one that is real (as said above).

ConditionalExpression[1/2 (1 + (2 I \[Pi] C[1])/Log[3] + Log[27]/Log[3]), C[1] \[Element] Integers]

This basically means that x can be the above expression for every choice of C[1] as an integer. C[1] = 0 corresponds to x = 2. The only Real solution...
POSTED BY: Sander Huisman
Ray Lawicki
Ray Lawicki, COLLEGE
Posted 10 years ago

thanks for the solution

POSTED BY: Ray Lawicki

You may use also the function Reduce[]:

In[1]:= Reduce[3^(2 x - 1) == 27, x, Integers]
        Reduce[3^(2 x - 1) == 27, x, Reals]

Out[1]= x == 2
Out[2]= x == 2
POSTED BY: Valeriu Ungureanu

Try Solve[3^(2x-1)==27,x,Integers] or Solve[3^(2x-1)==27,x,Reals], it gives you 2, which should be the answer.
POSTED BY: Syed Huq
Ray Lawicki
Ray Lawicki, COLLEGE
Posted 10 years ago

thanks for the solution i did not realise that restricting the domain to real was the trick
POSTED BY: Ray Lawicki
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