algebrasheet2 solution

algebrasheet2 solution - 1 Solutions to the Second Algebra...

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Unformatted text preview: 1 Solutions to the Second Algebra Worksheet 1.1 Solve e 2 x- 2 e x- 3 = 0 for x . This is a quadratic problem. Let z = e x . Then, we have that: z 2- 2 z- 3 = 0 ⇒ ( z- 3)( z + 1) = 0 ⇒ z = 3 or z =- 1 ⇒ e x = 3 or e x =- 1 ⇒ x = ln(3) or x = ln(- 1) ⇒ x = ln(3) . 1.2 Solve 1 2 x- 8 x 1 / 3 = 0 for x . This is a factoring problem. 1 2 x- 8 x 1 / 3 = 0 ⇒ x 1 / 3 1 2 x 2 / 3- 8 ¶ = 0 ⇒ x 1 / 3 = 0 or 1 2 x 2 / 3- 8 = 0 ⇒ x = 0 or x 2 / 3 = 16 ⇒ x = 0 or x = 16 3 / 2 ⇒ x = 0 or x = 4 3 ⇒ x = 0 or x = 64 . 1.3 Solve 2 x = 3 e 4 x for x . With problems involving things being raised to the x power it is a good idea to try to use logs. Just be sure to use the properties correctly!! 2 x = 3 e 4 x ⇒ ln(2 x ) = ln(3 e 4 x ) ⇒ x ln(2) = ln(3) + ln( e 4 x ) ⇒ x ln(2) = ln(3) + 4 x ⇒ x ln(2)- 4 x = ln(3) ⇒ x (ln(2)- 4) = ln(3) ⇒ x = ln(3) ln(2)- 4 . 1.4 Solve 1- 1 1+ x = k for x . Whenever what you are solving for appears in the denominator of a fraction it is a good idea to “clear” the fraction by multiplying by the denominator. 1- 1 1 + x = k ⇒ (1 + x ) 1- 1 1 + x ¶ = k (1 + x ) ⇒ 1 + x- 1 = k + kx ⇒ x = k + kx ⇒ x- kx = k ⇒ x (1- k ) = k ⇒ x = k 1- k 1.5 Solve e x- 2 = k 4 x for x . e x- 2 = k 4 x ln( e x- 2 ) = ln( k 4 x ) x- 2 = ln( k ) + ln(4 x ) x- 2 = ln( k ) + x ln(4) x- x ln(4) = ln( k ) + 2 x (1- ln(4) = ln( k ) + 2 x = ln( k ) + 2 1- ln(4) . 1.6 Solve 4 x 3 e kx- 16 xe kx = 0 for x . 4 x 3 e kx- 16 xe kx = 0 4 xe kx ( x 2- 4) = 0 4 xe kx ( x- 2)( x + 2) = 0 4 x = 0 or or e kx = 0 or x- 2 = 0 or x + 2 = 0 ⇒ x = 0 or or kx = ln(0) or x = 2 or x =- 2 ⇒ x = 0 or x = 2 or x =- 2 . 1.7 Solve 2 ln(5 x ) = ln( x + 2) for x ....
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This note was uploaded on 03/12/2012 for the course MATH 161 taught by Professor C during the Fall '07 term at Arizona.

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algebrasheet2 solution - 1 Solutions to the Second Algebra...

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