# 指数と対数の独自導出 / Original derivation of exponents and logarithms 作成・実読確認日 / Checked: 2026-10-06. Author: /root/article_author_oct6. 対象は実数の指数・対数、三例題の計算・全選択肢比較、読解練習。外部の科学事実・歴史・学習効果・検索成果を主張しない。外部一次本文を取得した記録でもない。 ## 1. 整数指数と逆向きの問い / Integer powers and the inverse question a>0、正の整数nに対しa^nはn個のaの積。積を並べるとa^m a^n=a^(m+n)。a^0=1、a^(-n)=1/a^nと定めれば、整数の指数法則を保つ。負の指数は負の値ではなく逆数を表す。a^(1/n)はaの正のn乗根である。実数の指数へ拡張したとき、a>0,a≠1なら指数と正の値は一対一に対応する。以下の付録はこの拡張と法則を自作の式で基礎付ける。 For positive integer n, a^n is the product of n copies of a. Juxtaposing products gives the addition rule. Setting a^0=1 and a^(-n)=1/a^n preserves that rule. A negative exponent gives a reciprocal, not a negative value. Positive roots give rational powers; the appendix constructs real powers with the same laws. log_a b=t means a^t=b, with a>0,a≠1,b>0. A base above one increases with t; a base between zero and one decreases. Both are one-to-one and cover all positive values. Base one is constant and cannot give a unique inverse. Zero and negative arguments are outside this real logarithm. log_a 1=0 follows from a^0=1; an argument of one is allowed. ## 2. 積の法則 / Product rule x>0,y>0としu=log_a x,v=log_a y。x=a^u,y=a^vなのでxy=a^(u+v)。逆向きに読むとlog_a(xy)=u+v=log_a x+log_a y。条件a>0,a≠1も保持。指数の足し算から生まれる法則で、log_a(x+y)の式ではない。 Let u=log_a x and v=log_a y. Then xy=a^u a^v=a^(u+v); the exponent representing the product is u+v. This works for positive arguments and a valid real base, including a base below one. ## 3. 底変換 / Change of base, without assuming the power rule a,b,c>0,a≠1,c≠1。t=log_a b、r=log_c a、s=log_c bと置く。a=c^r,b=c^s,a^t=bなので(c^r)^t=c^s。実指数の累乗法則からc^(rt)=c^s。一対一性によりrt=s。a≠1ならr≠0（c^0=1）なのでt=s/r。従ってlog_a b=log_c b/log_c a。分子は求める数bの対数、分母は元の底aの対数。b=1ではs=0となり答えは0であり、逆比は分母0になってしまう。 With t,r,s defined above, (c^r)^t=c^s implies rt=s. Divide by r, which is nonzero because a≠1. The target argument belongs in the numerator and the original base in the denominator. A valid new base c is required too. This derives the formula from exponent laws rather than presupposing the logarithmic power rule. ## 4. 例題全選択肢 / All choices in the existing examples ma-0126: 16=2^4. Base two is one-to-one, so x+1=4 and x=3 (b). The four existing candidates x=2,3,4,5 yield 2^3=8,2^4=16,2^5=32,2^6=64 respectively; only b satisfies the equation. Distinguish x from the whole exponent x+1. The exact JA/EN question, explanation, revision, source objects and approval/hash are in approved-examples.json. No illustration exists in either question object; this absence is preserved. ma-0044: u=log_10 x,v=log_10 y are arbitrary real exponents because x,y can be any positive real numbers. Required expression is u+v (d). Option a is uv; option b is u-v; option c is u/v and is undefined for y=1. These are not equal to u+v for all u,v. As an algebraic countercheck, at u=v=1 the required value is2, while the three rejected expressions are1,0,1. This is a proof countercheck, not a new quiz, translated example or added bank item. The rendered article only uses the original symbolic question. ma-0045: required expression is s/r (a), where r≠0. Option b is r/s; it is undefined for allowed b=1, and reverses the ratio. Option c is r+s, option d is rs, neither an identity for s/r. An algebraic countercheck at r=s=2 gives1 for the required expression and4 for c/d. Some wrong choices can coincide at special inputs; that does not establish a formula for all permitted values. The exact existing option strings are preserved. ## 5. 実数指数を自己完結させる付録 / A self-contained real-power construction This appendix supplies the mathematical basis; its series construction is not required reading for the article or a new quiz example. Define E(t)=sum from n=0 to infinity of t^n/n! for real t. The absolute ratio |t|/(n+1) eventually falls below1/2, giving a geometric tail and absolute convergence. On every bounded interval |t|≤M, the same tail estimate holds uniformly; polynomials are continuous and the uniform tail tends to zero, so E is continuous. Absolute convergence permits regrouping the double product. The coefficient with total degree k in E(u)E(v) is sum over n=0..k of u^n v^(k-n)/(n!(k-n)!)=(u+v)^k/k! by the finite binomial formula. Thus E(u)E(v)=E(u+v), E(0)=1 and E(t)E(-t)=1. Also E(t)=E(t/2)^2>0, since E(t/2) cannot be zero. For h>0, the positive terms give E(h)>1, and E(t+h)=E(t)E(h)>E(t). Thus E is strictly increasing. For t≥0, E(t)≥1+t tends to infinity; E(-t)=1/E(t) tends to zero. Continuity and the intermediate value property supply a unique real L(b) with E(L(b))=b for every b>0. L(1)=0, and strict increase gives L(a)>0 for a>1 and L(a)<0 for00. Then a^1=a, a^0=1, a^(u+v)=a^u a^v. Integer powers therefore equal repeated products, and negative powers equal reciprocals. For positive integer n, (a^(1/n))^n=a, so it is the positive nth root; uniqueness follows from strict increase of the product power on positive values. These definitions agree with rational powers. For any positive a and real r, E(rL(a))=a^r implies L(a^r)=rL(a) by the uniqueness of L. Consequently (a^r)^t=E(tL(a^r))=E(trL(a))=a^(rt). If a≠1, L(a)≠0. The map t↦a^t is continuous, strictly increasing for a>1 and strictly decreasing for0