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Metamath Proof Explorer

Theorem rxp112d

Description: Real exponentiation is one-to-one with respect to the second argument. (TODO: Note that the base C must be positive since -u C ^ A is C ^ A x.e ^ i _pi A , so in the negative case A = B + 2 k ). (Contributed by SN, 25-Apr-2025)

Ref Expression
Hypotheses rxp112d.c ⊢ ( 𝜑 → 𝐶 ∈ ℝ+ )
rxp112d.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
rxp112d.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
rxp112d.1 ⊢ ( 𝜑 → 𝐶 ≠ 1 )
rxp112d.2 ⊢ ( 𝜑 → ( 𝐶 ↑𝑐 𝐴 ) = ( 𝐶 ↑𝑐 𝐵 ) )
Assertion rxp112d ( 𝜑 → 𝐴 = 𝐵 )

Proof

Step Hyp Ref Expression
1 rxp112d.c ⊢ ( 𝜑 → 𝐶 ∈ ℝ+ )
2 rxp112d.a ⊢ ( 𝜑 → 𝐴 ∈ ℝ )
3 rxp112d.b ⊢ ( 𝜑 → 𝐵 ∈ ℝ )
4 rxp112d.1 ⊢ ( 𝜑 → 𝐶 ≠ 1 )
5 rxp112d.2 ⊢ ( 𝜑 → ( 𝐶 ↑𝑐 𝐴 ) = ( 𝐶 ↑𝑐 𝐵 ) )
6 2 recnd ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
7 3 recnd ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
8 1 relogcld ⊢ ( 𝜑 → ( log ‘ 𝐶 ) ∈ ℝ )
9 8 recnd ⊢ ( 𝜑 → ( log ‘ 𝐶 ) ∈ ℂ )
10 1 4 logne0d ⊢ ( 𝜑 → ( log ‘ 𝐶 ) ≠ 0 )
11 5 fveq2d ⊢ ( 𝜑 → ( log ‘ ( 𝐶 ↑𝑐 𝐴 ) ) = ( log ‘ ( 𝐶 ↑𝑐 𝐵 ) ) )
12 1 2 logcxpd ⊢ ( 𝜑 → ( log ‘ ( 𝐶 ↑𝑐 𝐴 ) ) = ( 𝐴 · ( log ‘ 𝐶 ) ) )
13 1 3 logcxpd ⊢ ( 𝜑 → ( log ‘ ( 𝐶 ↑𝑐 𝐵 ) ) = ( 𝐵 · ( log ‘ 𝐶 ) ) )
14 11 12 13 3eqtr3d ⊢ ( 𝜑 → ( 𝐴 · ( log ‘ 𝐶 ) ) = ( 𝐵 · ( log ‘ 𝐶 ) ) )
15 6 7 9 10 14 mulcan2ad ⊢ ( 𝜑 → 𝐴 = 𝐵 )