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Exponential & logarithmic equations #214
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@MomoDeve wanna be an assignee for this one?
a ^ f(x) + b ^ g(x) + ... + t = 0can be solved only if all functions are linear or can be converted to linear by substitution: for example,2^(x^2 + 1) + 2^(2*x^2 - 1) + 1 = 0can be solved, but2^(x + 1) + 2^(2*x^2 - 1) + 1 = 0cannot.- where are no exponential parsers. I have an idea how to solve such equations, but it requires a lot of dirty work just to determine if an equation is solvable and which substitution to choose
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on Aug 4, 2026 - added a commit that references this issue
on Aug 8, 2026 Measured on
masterata45a7256— this looks substantially done, including the general form the body asks for.So that
a ^ f(x) + b ^ g(x) + ... = 0could be solved. And, likewise, a logarithmic oneequation answer 2^x - 8{ 3 }2^x + 2^(-x) - 5/2{ 1, -1 }e^(2x) - 3e^x + 2{ 0, ln(2) }2^x + 3^x - 5{ 1 }— different bases, which is the general form2^(2x) - 5*2^x + 6{ 1, ln(3 ^ (1 / ln(2))) }4^x - 3*2^x + 2{ 0, 1 }ln(x) - 2{ e ^ 2 }ln(x)^2 - 3*ln(x) + 2{ e, e ^ 2 }log(2,x) + log(4,x) - 3{ 4 }@MomoDeve's analysis in the thread — that these are solvable where the exponents are linear or become linear under a substitution — is what the solver now does, and the mixed-base case works too.
Two things are imperfect rather than missing, and I would suggest they are separate and smaller than this issue:
2^(2x) - 5*2^x + 6answersln(3 ^ (1 / ln(2)))wherelog(2, 3)says the same thing far more plainly. Correct, ugly.ln(x) + ln(x+1)answers0.6180339887...— a decimal, where the exact root is(sqrt(5) - 1) / 2. That is a loss of exactness rather than a formatting complaint, and of the two it is the one I would actually file.
Happy to open (2) as its own issue if you want it tracked; otherwise this one looks closeable.
Measured on
masterat6b93b401(2.3.0), .NET 10. Both halves of this issue are answered —a^f(x) + b^g(x) + … = 0and the logarithmic mirror of it:2^x - 8 -> { 3 } 2^x + 3^x - 5 -> { 1 } 2^(2x) - 5*2^x + 4 -> { 0, 2 } e^x - e^(-x) - 1 -> { ln((1 - sqrt(5))/2), ln((1 + sqrt(5))/2) } 5^x - 7^x -> { 0 } ln(x) + ln(x - 1) - ln(6) -> { 3 } log(2, x) + log(2, x - 2) - 3 -> { 4 } ln(x)^2 - 3*ln(x) + 2 -> { e, e^2 }The implementation is
Sources/AngouriMath/Functions/Continuous/Solvers/EquationSolver/ExponentialSolver.cs(SolveLinear,SolveMultiplicative,GetConstantOutOfLogarithm) together with the fresh-variable substitution inAnalyticalEquationSolver, and it is pinned bySources/Tests/UnitTests/Algebra/AcceptedProposalsAlreadyImplementedTest.cs— so it is covered rather than merely working today. #246, the logarithmic half filed separately, is already closed.One residual, filed separately as #1007 rather than left in this thread: a mixed-base exponential comes back with a
doublewhere an exact form exists —3^(x+1) - 2^(x-1)givesln(0.04674569822628630438…^(1/ln(2)))where the answer is-ln(6)/ln(3/2), agreeing to 17 significant figures and then diverging. That is a precision defect in one branch, not this proposal being unimplemented.Closing on the measurement, per #746 item 1. This is one of the "open issues that are already done" that a sweep keeps turning up — the third today.
So that
a ^ f(x) + b ^ g(x) + ... = 0could be solved. And, likewise, a logarithmic one