# Solving Exponencial fuction is not returning the right answer

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Rodrigo Toledo on 2 Apr 2021
Commented: William Rose on 4 Apr 2021
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 on 2 Apr 2021
Edited: John D'Errico on 2 Apr 2021
Yes, you THINK the true answer is 3. But it is not.
format long g
log(2653/2500) / log(51/50)
ans =
2.99961931278214
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?
12000*sym('1.02')^3
ans =
12734.496
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 =
3
fval =
0.0960000000013679
exitflag =
1
Here MATLAB was able to find an integer solution.
##### 2 CommentsShowHide 1 older comment
William Rose on 3 Apr 2021
You're elcome Rodrigo.

William Rose on 2 Apr 2021
Algebra:
x=log(12734/12000)/log(1.02)
>> x=log(12734/12000)/log(1.02)
x =
2.9980
>>
William Rose on 4 Apr 2021
@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 ).