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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 ) # S EsCbRi^E'T T o R e G e n e ra l C o u n se l F ro m C h a rlo tte ($ ) 2 7 8 H Q C 1 2 2 9 7 3 6 V IO , 0 2 /2 1 /2 0 0 7 b l b 6 b 7 C. ( ( U 1 # X ¦ , 5 & A A U X í ô l î Ã # X W ~ ð õ ñ õ î ò r ó õ r ó ì U 4 !.

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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. āI āI 㔭 I ̊ ^ J R e X g C x g I ܂ Ǒ u H ̎ԉ Ձi Ⴈ 񂳂 j v ֐ I ̊ ^ J R e X g C x g I ܂ Ǒ u H ̎ԉ Ձv G g ₢ 킹 X V. A a Ma a , I a a B T a Ja M , G a a M E , I c.

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2 (a) Deﬁne uniform continuity on R for a function f R → R (b) Suppose that f,g R → R are uniformly continuous on R (i) Prove that f g is uniformly continuous on R (ii) Give an example to show that fg need not be uniformly continuous on R Solution • (a) A function f R → R is uniformly continuous if for every ϵ > 0 there exists δ > 0 such that f(x)−f(y) < ϵ for all x. `V` dghVX^hr \^cr, `dchfda^fibiä eadhrä, ^ Xd_h^ X cdXiä \^cr, YZ XaVZqmghXih Xåhd_ @ik D Xdh dZcV\Zq, X 1997 YdZi, å dhmha^Xd igaqnVa. B Í ç ® m 4 å ' ) j ' r s k % j @ Ò é ' ) i 6 ¹ q À 5 = ã v Ô q v w 4 * % £ b > "Ç } a "È ¸ ö ô j t b Ú ^ ) ( 7 ¼ ) ( * g i ö r ° Ã j Ã ¼ ) ( * i ÷ c "É æ # Ò é Ã ¼ é ^ a m Ö a 4 b ÿ 7 ç c 7\hvxex#\dkrr fr ms ì g Ç x x.

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