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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. I K X s h } X ^ f C f C g Y r v OMEGA v ( Speed Master DayDate ) u b N/OM3250 V i/ l C/ s/ Ԃ / / h / @ B / / N m O t/ X C X/ ^ x g/3250/ V o I X X x. N u c Y @ c A C x g u z K ΐV ԉΑ ƐM B ӍՁI v y z @ @ @ @ 쌧 z K S x m x m Ԓn.

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