Unexpected result of equation solving in Wolfram|Alpha

Questions
,
1

Good day for all!

There is an incorrect result for equation solving

over the reals.
There must be only one corrent solution x=1, but WA and WM (Reduce / Solve) also provides x=0 which is incorrect.
Why doesn’t WA/WM check the validness of its results?

2

I suppose that Mathematica makes the following simplification
In[103]:= Simplify[(8^x - 2^x)/(6^x - 3^x)]

Out[103]= (2/3)^x (1 + 2^x)

and then does not check back the original equation. This is a well-known issue:

In[104]:= Solve[(x^2 - 1)/(x - 1) == 2, x]

Out[104]= {{x -> 1}}

However, just plot the function

Plot[(8^x - 2^x)/(6^x - 3^x) - 2, {x, -1, 2}]

and you will see that Mathematica’s answer is not so incorrect in some sense.

FunctionDomain checks back on the original form of the function. Try this:

Solve[(8^x - 2^x)/(6^x - 3^x) == 2 && 
  FunctionDomain[(8^x - 2^x)/(6^x - 3^x), x], x, Reals]
3

Thanks for your answer!

But nevertheless, WM/WA should check back the original equation automatically without extra functions such as FunctionDomain when the original and simplified equations are not equivalent.

4

I am not familiar with the workings of W|A in this case, but Solve will handle removable singularities by removing them.

5

Well, if I understood your right, it’s not true: as of 13.0.1 W|M Solve doesn’t remove singular solutions.
Look at the example above with Solve[(x^2 - 1)/(x - 1) == 2, x].

Or did you mean it will be fixed in the future versions of W|M?

6

I didn’t claim solutions at singularities would be removed. I stated that the singularities themselves would be removed, that is, not be regarded as singularities (that’s what it means to remove a removable singularity). Stated in terms of the most recent example, in the field of rational functions (x^2-1)/(x-1) is equivalent to (x+1).

This is not regarded as broken by the way. Wolfram|Alpha is more in tune with what is taught in high schools, and will attempt to remove singular solutions. So that particular example will have an empty solution set. Different philosophy.

7

If Mathematica removed all removable singularities I would expect this to give x->0:

Solve[Sin[x]/x == 1 && Abs[x] < 1, x]

just as I get for this:

Solve[Sinc[x] == 1 && Abs[x] < 1, x]

Maybe the full story is more complicated.

8

You are correct, it is only removing the algebraic ones, that is, those that arise from rational functions. It is not removing more general ones from meromorphic functions.