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An equation linear in logarithms is solved by exponentiating - #2316
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An equation in two or more logarithms of x had no route of its own, and came to Newton's
method, which answered it in decimals and left out the roots it did not reach:
ln(x) + ln(x + 1) = 0 was { 0.618... } and ln(1 + 3x^2) - ln(4x) = 1 was { 3.53 }, without
0.0944.
k_1 log_b(P_1) + ... + k_n log_b(P_n) = c, for polynomials P_i, rational k_i and one base,
holds only where P_1^(k_1 L) ... P_n^(k_n L) = b^(c L), L being the common denominator of
the k_i: a polynomial equation, whose roots are found exactly. Each is put back into the
equation as written under the principal logarithm and kept only where it holds, since over
the complex numbers ln(u) + ln(v) is ln(u v) only up to 2 pi i: at (-1 - sqrt(5))/2 both
logarithms of ln(x) + ln(x + 1) carry i pi, and the sum is 2 pi i. Over the reals each
argument is to be positive as well. A residual too small to tell from rounding, or a candidate that is not a root of the
polynomial, leaves the equation to the routes after, since the roots the polynomial solver
missed would be left out: 3x^3 + 9x^2 + 9x + 1, a shifted cube, comes back as -1 three
times until #2315.
On 400 random equations of this kind, checked against a Newton search under the principal
logarithm, master answers 292, leaves 108 unsolved and misses 25 roots in 22 of its
answers; this answers all 400, none unsound and none missing a root.
Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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Files 311 312 +1
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- Misses 4877 4890 +13
- Partials 4141 4154 +13
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…entiating Co-Authored-By: Claude Opus 5.5 (1M context) <noreply@anthropic.com>
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An equation in two or more logarithms of
xhad no route of its own. It came to Newton's method, which answered in decimals and left out any root it didn't reach.k_1 log_b(P_1) + … + k_n log_b(P_n) = c, for polynomialsP_i, rationalk_iand one base, holds only whereP_1^(k_1 L) … P_n^(k_n L) = b^(c L), withLthe common denominator of thek_i. That is a polynomial equation, and its roots are found exactly. Each one is put back into the equation as written, under the principal logarithm, and kept only where it holds, since over the complex numbersln(u) + ln(v)isln(u v)only up to2 pi i. Over the reals each argument must also be positive.Solve("x")ln(x) + ln(x + 1) = 0{ 0.6180339887498947… }{ (-1 + sqrt(5)) / 2 }ln(1 + 3x^2) - ln(4x) = 1{ 3.529945609977906… }(4e ± sqrt(16e^2 - 12)) / 6ln(-1 - 3x^2) - ln(4x) = 1{ -3.529945609977908… }, and on master… - 1.3E-15ii picancelln(x^2 + 1) - ln(x) = 1(e ± sqrt(e^2 - 4)) / 2ln(x) + ln(x - 1) = ln(6){ 3 }{ 3 }; at-2the sum isln(6) + 2 pi i3x^3 + 9x^2 + 9x + 1is one, since the polynomial solver answers it with a non-root until A cube shifted is solved #2315.BREAKING-CHANGES.mdhas an entry.No issue was filed.
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