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This is part of a long question im doing I have just come from e^y = e^x c to get to y = ln(e^x c) so dont want to go backwards, (e both sides) I recall (ln) rules are a bit wierd so im sorta stuck I final got y = x ln(c), But i dono if i can ln(e^x c) = ln(e^x) ln(c) Can some1 check this?. Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more. Z y W 𒆐S Ƃ ` ̑ A h o C U B l s V Œn ň Ō ʓI ȃz y W Ă ܂ B u A V X g y y v o R g ܂ Ȃ ł A A V X g y y ͌ l Ƃł B.

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E x= ex > 0 ⇒ E(x) = e is concave up, increasing, and positive Proof Since E(x) = ex is the inverse of L(x) = lnx, then with y = ex, d dx ex = E0(x) = 1 L0(y) = 1 (lny)0 = 1 1 y = y = ex First, for m = 1, it is true Next, assume that it is true for k, then d k1 dxk1 ex = d dx d dxk ex = d dx (ex) = ex By the axiom of induction, it is. C q ̃ H C X G i W N V b v L ̏ڂ e 11 N3 8 E22 i ΁j. Nature voice e l c ` { c x e @ @ @ r y 𒆐s ɍ y ̃p ̎ 舵 y сa q o y.

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E x= ex > 0 ⇒ E(x) = e is concave up, increasing, and positive Proof Since E(x) = ex is the inverse of L(x) = lnx, then with y = ex, d dx ex = E0(x) = 1 L0(y) = 1 (lny)0 = 1 1 y = y = ex First, for m = 1, it is true Next, assume that it is true for k, then d k1 dxk1 ex = d dx d dxk ex = d dx (ex) = ex By the axiom of induction, it is. { S ̃^ C X g A ^ C T u g N v Ɨ 悤 I n A X ܖ A E ޗ E X ^ b t ł ł ܂ B Tweet. W ł ǂ ł C g Q I x R Z v g ɁA5 { p E d 485cm ƃR p N g Ɍg !.

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