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Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N. U y v( o t) A y R T g Ēv ܂ B A c A q 炲 ҂܂Ŋy ߂ B c t E ۈ牀 v O 1 E w Z Z u ӏ܉ ͍ N ł ߂悤 Ǝv Ă ̂ v. Smith, R Snodgrass, S F Terrell, D L Thmnas, B J Clit' kiflc cam x if' RIFLE TEAM LETTERMEN Back row to nght Col Raymond Orr George T Baker Kenneth L Flelder C W Levmgston V K Cameron Thomas E Werlm Bemgno Leal Col Stanley Y Kennedy Marcel Kenty Vlctor D Everett Not pictured Mrlton L Bell Walks H Eckhoff john T Hom HEARST TROPHY MATCH.

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