Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Change of base and inversion of the positive-base real exponential

Statement

If b>0b>0, b≠1b\ne1, and x>0x>0, then log⁡bx=log⁡xlog⁡b,blog⁡bx=x,log⁡b(bu)=u(u∈R).\log_bx=\frac{\log x}{\log b},\qquad b^{\log_bx}=x,\qquad \log_b(b^u)=u\quad(u\in\mathbb R). In particular, log⁡bx=log⁡cx/log⁡cb\log_bx=\log_cx/\log_cb for every second base c>0c>0, c≠1c\ne1.

Facts & Assumptions

Given: b,c>0b,c>0 with b,c≠1b,c\ne1, x>0x>0, and u∈Ru\in\mathbb R.

[L1]

log⁡bx=log⁡x/log⁡b\log_bx=\log x/\log b, with log⁡b≠0\log b\ne0 (The logarithm to a positive base other than one).

[L2]

bv=exp⁡(vlog⁡b)b^v=\exp(v\log b) and the real-power laws hold for positive bases (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents).

[L3]

log⁡(exp⁡v)=v\log(\exp v)=v and exp⁡(log⁡x)=x\exp(\log x)=x for x>0x>0 (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

technique · direct
1.1

By [L1] and [L2], blog⁡bx=exp⁡((log⁡x/log⁡b)log⁡b)=exp⁡(log⁡x)=xb^{\log_bx}=\exp((\log x/\log b)\log b)=\exp(\log x)=x.

L1L2L3
1.2

Likewise log⁡b(bu)=log⁡(exp⁡(ulog⁡b))/log⁡b=u\log_b(b^u)=\log(\exp(u\log b))/\log b=u.

L1L2L3
2.1

Dividing log⁡x\log x first by log⁡c\log c and then by log⁡b/log⁡c\log b/\log c gives log⁡bx=log⁡cx/log⁡cb\log_bx=\log_cx/\log_cb.

L1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 37 results over 13 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources