Explanation Given: The equation e 2 x − 6 e x + 8 = 0 , which is in the quadratic form. The objective is to find the solution for the equation. The problem can be approached in two steps, First we can treat e x as a variable y which converts the given equation into a quadratic equation in y . The roots of the quadratic equation can be found by using any of the methods to solve a typical quadratic equation. In the next part we can equate each root to e x and obtain the values for x . Formula used: Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values. 1) log b ( m n ) = log b ( m ) + log n ( n ) 2) log b ( m n ) = log b ( m ) − log n ( n ) 3) log b ( m n ) = n ⋅ log b ( m ) 4) log b ( m ) = n ⇒ m = b n ; log b b = 1 The natural log is the inverse of e . 5) ln ( m ) = n ⇒ m = e n 6) e ln ( m ) = m The Latin name for the natural log is logarithmus naturali , which gives the abbreviation for natural log as ln . The natural log of a number is the log of the number to the base e . ln ( m ) = log e ( m ) Most commercial calculators can compute natural log of numbers and log to the base 10. The standard quadratic equation has the form a x 2 + b x + c = 0 , the roots of the equation are given by the formulae x 1 = − b + D 2 a and x 2 = − b − D 2 a , where the discriminant D is given by the formula D = b 2 − 4 a c Calculation: The problem is solved in two steps. In the first step, substitute e x = y in the given equation e 2 x − 6 e x + 8 = 0 , to get y 2 − 6 y + 8 = 0 and solve for y . Description y 2 − 6 y + 8 = 0 Step 1: The equation is in the standard form a x 2 + b x + c = 0 in y , with a = 1 , b = − 6 and c = 8

College Algebra with Modeling & Visualization (6th Edition)

6th Edition

ISBN: 9780134441795

Author: Rockswold

Publisher: PEARSON

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Chapter 5.6, Problem 77E

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The solution of the given equation which is in quadratic form.

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Chapter 5.6, Problem 76E

Chapter 5.6, Problem 78E

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