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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 ∈ NË t s Ô ù U _ M b {Ù g w ý ` M ± z _ m Z h è µ Ä å ï ® 4 ) " 3 *¯ ¦ N ã ( q § Å w b ;H x Ô p å ï ½ ~ Ã Æ » Ü Þ è µ Ä å ï w w b Ë 0 ' ;

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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 ∈ NSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more³ >/ '¨>/ ³ " ) \ ^ 1* !l \ K Z '¨ G g1 7T1 (3û 4E ò ì _0¿ I S & Q#Ý K S >0 ³ _7 K Z b Æ _ > E w

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