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Solving Exponencial fuction is not returning the right answer

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Rodrigo Toledo
Rodrigo Toledo 2021 年 4 月 2 日
コメント済み: William Rose 2021 年 4 月 4 日
How do i solve this exponencial:
12734.40 == 12000 * 1.02^x
i am doing:
solve(12734.4 == 12000*1.02^x,x
and getting log(2653/2500) / log(51/50)
the answer shoud be a 3.


John D'Errico
John D'Errico 2021 年 4 月 2 日
編集済み: John D'Errico 2021 年 4 月 2 日
Yes, you THINK the true answer is 3. But it is not.
format long g
log(2653/2500) / log(51/50)
ans =
And that is effectively the exact answer (to 16 significant digits) to the problem you posed. You could have made MATLAB convert that number to a floating point number, using either vpa or double.
What would the true left hand side have been, if 3 had been the correct result?
ans = 
So you had 12734.40 in the equation you tried to solve. Should MATLAB have solved a different problem than the one you posed? Surely not! MATLAB cannot know that you only had approximate coefficients, and that you expected an integer result. At least not unless you tell it to look for an approximate solution using some more appropriate solution method.
For example, if I use a solver that can constrain the unknown to be an integer, minimizing the absolute error, I get this:
fun = @(x) 12734.40 - 12000*1.02.^x;
lb = 0;
[xval,fval,exitflag] = ga(@(x) abs(fun(x)),1,[],[],[],[],lb,[],[],1)
Optimization terminated: average change in the penalty fitness value less than options.FunctionTolerance and constraint violation is less than options.ConstraintTolerance.
xval =
fval =
exitflag =
Here MATLAB was able to find an integer solution.
  2 件のコメント
William Rose
William Rose 2021 年 4 月 3 日
You're elcome Rodrigo.


その他の回答 (1 件)

William Rose
William Rose 2021 年 4 月 2 日
>> x=log(12734/12000)/log(1.02)
x =
  4 件のコメント
William Rose
William Rose 2021 年 4 月 4 日
@Rodrigo Toledo, I am not familiar with Matlab's symbolic math routines. I don't know if you can instruct Matlab to restrict the solution space to (or restrict it to ).



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