simplify returns incorrect result

[devmotion]

I ran into the following issue on Symbolics#master:

julia> using Symbolics

julia> ex = (1.1540262529490819e-7(5.011872336272725e-8 + a)*(10^(-x))) / (2.511886431509582e-15((a + 10^(-x))^2)) + (-2.302585092994046(5.011872336272725e-8 + a)*(10^(-x))) / (5.011872336272725e-8(a + 10^(-x)));

julia> Symbolics.simplify(ex)
0

even though the expression is clearly not zero. As an example, compare

julia> Symbolics.substitute(ex, Dict(x => 1.0, a => 1.0))
-379691.03250567336

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A bit longer story: I encountered this problem when using Symbolics.hessian(...; simplify=true). I got

julia> using Symbolics

julia> @variables t Ka ktr ph(t) tr1(t);

julia> D = Differential(t);

julia> ex = 10^(-ph)/(10^(-ph) + Ka) / (10^(-7.3) / (10^(-7.3) + Ka)) - ktr * tr1;

julia> Symbolics.hessian(ex, [ph, tr1]; simplify=true)
2×2 Matrix{Num}:
 0  0
 0  0

even though the second derivative of ex with respect to ph should be non-zero

julia> Symbolics.hessian(ex, [ph, tr1])
2×2 Matrix{Num}:
 (-2.65724e-7(5.01187e-8 + Ka)*(10^(-2ph(t)))) / (2.51189e-15((Ka + 10^(-ph(t)))^2)) + (5.3019(5.01187e-8 + Ka)*(10^(-ph(t)))) / (5.01187e-8(Ka + 10^(-ph(t)))) + 1.15403e-7(10^(-ph(t)))*((-2.30259(5.01187e-8 + Ka)*(10^(-ph(t)))) / (2.51189e-15((Ka + 10^(-ph(t)))^2))) + 1.15403e-7((-2.30259(5.01187e-8 + Ka)*(10^(-ph(t)))) / (2.51189e-15((Ka + 10^(-ph(t)))^2)) + 1.15677e-14(Ka + 10^(-ph(t)))*(10^(-ph(t)))*(((5.01187e-8 + Ka)*(10^(-ph(t)))) / (6.30957e-30((Ka + 10^(-ph(t)))^4))))*(10^(-ph(t)))  …  0
                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                              0     0

The example above is actually the slightly simpler case

julia> Symbolics.derivative(ex, ph; simplify=true)
0

julia> Symbolics.derivative(ex, ph)
(-2.302585092994046(5.011872336272725e-8 + Ka)*(10^(-ph(t)))) / (5.011872336272725e-8(Ka + 10^(-ph(t)))) + 1.1540262529490819e-7(((5.011872336272725e-8 + Ka)*(10^(-ph(t)))) / (2.511886431509582e-15((Ka + 10^(-ph(t)))^2)))*(10^(-ph(t)))

[ChrisRackauckas]

Ooof that's not good. @shashi do you have an idea for where to start looking for this?