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A constant raised to a multiple of a logarithm of itself is a power of the antilogarithm - #1430
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…f the antilogarithm c ^ (k * log(c, b)) is b ^ k: c ^ z is exp(z ln c) and log(c, b) is ln(b) / ln(c) on the principal branch, so the two ln(c) cancel and exp(k ln b) is left, which is what b ^ k means for every complex b other than 0. The k = 1 case folded since #1428 and this shape did not, and it is the shape the integrator's exponential substitution writes its antiderivatives in: e ^ (-3/2 * ln(u)) stood in answers that were otherwise algebraic. Three arms by what can be said of the base, as for the k = 1 rules, in both spellings: a number other than 0 and 1, a constant, and a symbol under the logarithm's own domain condition. The multiplier may sit on either side of the logarithm. The constant arm is written with a pattern on the right so it is data in both directions. The guide's counts and the pinned lists move: 332 rules, 303 names, 329 identities, 146 conditional, 126 Collects, 34 Expands, 30 subsumptions (constant before symbol), 963 subsumption claims, 36 two-sided rules of which 32 have a direction, and the longest name is nineteen words. BREAKING-CHANGES.md rows measured on a v2.5.0 build. #994 #718 Co-Authored-By: Claude Opus 5 (1M context) <noreply@anthropic.com> Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura
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…eal (#1431) The linear-radical substitution u = (a x + b)^(1/q) and the general substitution for an even root both take a power of u out of a radical on the strength of u being a non-negative real, which it is only where the radical is. Beyond the radicand's zero u is imaginary and the integrand can still be real: Rubi's x sqrt(c - a c x)/e^(3 atanh(a x)) above a x = 1 is the product of two imaginary factors, and the answer built for a real u was 14% off its derivative there and shipped as an answer (family 7, one wrong after #1430 made the row reachable). An answer that used the step now carries provided a x + b >= 0: NaN beyond it, which is no claim, and unchanged where it holds. An answer the step did not touch is as it was. #718 Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5 (1M context) <noreply@anthropic.com>
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…r it is (#1434) e^(k ln(q)) is q^k, the definition of the principal power for every complex q other than zero, and that is the spelling the parser gives every inverse hyperbolic function: acoth(a x) is 1/2 ln((a x + 1)/(a x - 1)). So e^acoth(a x) x^3 arrived as an exponential of a logarithm, which no exponential rule reads, and is x^3 sqrt((a x + 1)/(a x - 1)), a root of a quotient of linears, which the radical substitution answers. The simplifier folds the shape since #1430; the integrand is not simplified before the rules see it, so the fold is a route of its own, asked as a question of its own so that the closed rules meet the folded integrand at the top. The multiplier is gathered from the whole product in the exponent, since e^(3 acoth(a x)) arrives as e^(3 (1/2 ln q)). Rubi's 7.4.2 (exponentials of the inverse hyperbolic cotangent): of fifteen rows the probe declined, seven are answered and verified. #718 Claude-Session: https://claude.ai/code/session_012sonx8iAspMiwRwokT1Ura Co-authored-by: Claude Opus 5 (1M context) <noreply@anthropic.com>
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The k = 1 case,
c ^ log(c, b) = b, folds for every base since #1428; the same identity with a multiplier on the logarithm,c ^ (k * log(c, b)) = b ^ k, did not fold for any. It is the shape the integrator's exponential substitution writes its antiderivatives in —e ^ (-3/2 * ln(1 + 2 / (a x - 1)))stood in answers that were otherwise algebraic (the acoth rows of Rubi's family 7, #718).Why it is exact.
c ^ zisexp(z ln c)andlog(c, b)isln(b) / ln(c)on the principal branch, so the twoln(c)cancel andexp(k ln b)is left, which is whatb ^ kmeans for every complexbother than 0 — including the negative reals, wheree ^ (1/2 * ln(-3))isi sqrt(3), the principal(-3) ^ (1/2).PowerOfALogarithmMultipleTest.TheValueIsPreservedsamples both signs and a fractional multiplier.Three arms by what can be said of the base, as for the
k = 1rules, in both spellings (MatchedRules.csand thePowerswitch), with the multiplier on either side of the logarithm:Simplifya-number-raised-to-a-multiple-of-a-logarithm-of-itself-is-a-power-of-the-antilogarithm2 ^ (k * log(2, x))→x ^ k,10 ^ (3 * log(10, x))→x ^ 3a-constant-raised-to-a-multiple-of-a-logarithm-of-itself-is-a-power-of-the-antilogarithm— a pattern on the right, so it is data both wayse ^ (2 * ln(x))→x ^ 2,e ^ (-3/2 * ln(u))→u ^ (-3/2),pi ^ (k * log(pi, x))→x ^ ka-symbol-raised-to-a-multiple-of-a-logarithm-of-itself-is-a-power-of-the-antilogarithm-where-defineda ^ (k * log(a, x))→x ^ k provided not a = 0 and not a = 11 ^ (2 * log(1, x))is stillNaN.LogarithmSpellingsTest(per rule set, base 2 against base e) passes unchanged, since every arm reads both.The guide's counts and the pinned lists (
WritingARule.md,RuleAuthoringGuideTest,ReversibleRuleTest,RulePriorityTest,RuleMetadataTest,StepAsASentenceTest,AddressableRulesTest) move: 332 rules, 303 names, 329 identities, 146 conditional, 126 Collects, 34 Expands, 30 subsumptions (constant before symbol, as fork = 1), 963 subsumption claims, 36 two-sided rules of which 32 have a direction (the constant arm forgets which constant, the fourth of its kind), and the longest name is now nineteen words.BREAKING-CHANGES.md rows measured on a v2.5.0 build. Full suite 12006 passed, 0 failed; benchmark gate PASSED (SimplifyHard 171,811,504 B, +0.1% on the recorded baseline).
Part of #994 and #718.
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