Metamath Proof Explorer


Theorem abscxp

Description: Absolute value of a power, when the base is real. (Contributed by Mario Carneiro, 15-Sep-2014)

Ref Expression
Assertion abscxp ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴𝑐 𝐵 ) ) = ( 𝐴𝑐 ( ℜ ‘ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 simpr ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → 𝐵 ∈ ℂ )
2 relogcl ( 𝐴 ∈ ℝ+ → ( log ‘ 𝐴 ) ∈ ℝ )
3 2 recnd ( 𝐴 ∈ ℝ+ → ( log ‘ 𝐴 ) ∈ ℂ )
4 3 adantr ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( log ‘ 𝐴 ) ∈ ℂ )
5 1 4 mulcld ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( 𝐵 · ( log ‘ 𝐴 ) ) ∈ ℂ )
6 absef ( ( 𝐵 · ( log ‘ 𝐴 ) ) ∈ ℂ → ( abs ‘ ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) = ( exp ‘ ( ℜ ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) )
7 5 6 syl ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( abs ‘ ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) = ( exp ‘ ( ℜ ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) )
8 remul2 ( ( ( log ‘ 𝐴 ) ∈ ℝ ∧ 𝐵 ∈ ℂ ) → ( ℜ ‘ ( ( log ‘ 𝐴 ) · 𝐵 ) ) = ( ( log ‘ 𝐴 ) · ( ℜ ‘ 𝐵 ) ) )
9 2 8 sylan ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ℜ ‘ ( ( log ‘ 𝐴 ) · 𝐵 ) ) = ( ( log ‘ 𝐴 ) · ( ℜ ‘ 𝐵 ) ) )
10 1 4 mulcomd ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( 𝐵 · ( log ‘ 𝐴 ) ) = ( ( log ‘ 𝐴 ) · 𝐵 ) )
11 10 fveq2d ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ℜ ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) = ( ℜ ‘ ( ( log ‘ 𝐴 ) · 𝐵 ) ) )
12 recl ( 𝐵 ∈ ℂ → ( ℜ ‘ 𝐵 ) ∈ ℝ )
13 12 adantl ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ℜ ‘ 𝐵 ) ∈ ℝ )
14 13 recnd ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ℜ ‘ 𝐵 ) ∈ ℂ )
15 14 4 mulcomd ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) = ( ( log ‘ 𝐴 ) · ( ℜ ‘ 𝐵 ) ) )
16 9 11 15 3eqtr4d ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( ℜ ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) = ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) )
17 16 fveq2d ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( exp ‘ ( ℜ ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) = ( exp ‘ ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) ) )
18 7 17 eqtrd ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( abs ‘ ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) = ( exp ‘ ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) ) )
19 rpcn ( 𝐴 ∈ ℝ+𝐴 ∈ ℂ )
20 19 adantr ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → 𝐴 ∈ ℂ )
21 rpne0 ( 𝐴 ∈ ℝ+𝐴 ≠ 0 )
22 21 adantr ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → 𝐴 ≠ 0 )
23 cxpef ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝑐 𝐵 ) = ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) )
24 20 22 1 23 syl3anc ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( 𝐴𝑐 𝐵 ) = ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) )
25 24 fveq2d ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴𝑐 𝐵 ) ) = ( abs ‘ ( exp ‘ ( 𝐵 · ( log ‘ 𝐴 ) ) ) ) )
26 cxpef ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ∧ ( ℜ ‘ 𝐵 ) ∈ ℂ ) → ( 𝐴𝑐 ( ℜ ‘ 𝐵 ) ) = ( exp ‘ ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) ) )
27 20 22 14 26 syl3anc ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( 𝐴𝑐 ( ℜ ‘ 𝐵 ) ) = ( exp ‘ ( ( ℜ ‘ 𝐵 ) · ( log ‘ 𝐴 ) ) ) )
28 18 25 27 3eqtr4d ( ( 𝐴 ∈ ℝ+𝐵 ∈ ℂ ) → ( abs ‘ ( 𝐴𝑐 𝐵 ) ) = ( 𝐴𝑐 ( ℜ ‘ 𝐵 ) ) )