Stitz-Zeager_College_Algebra_e-book

3 and 64 in equation 63 we use p to denote the

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Unformatted text preview: 3. Since u = 5x , we have 5x = −2 or 5x = 3. Since 5x = −2 has no real solution, (Why not?) we focus on 5x = 3. Since it isn’t convenient to express 3 as a power of 5, we take natural logs and get ln (5x ) = ln(3) so that x ln(5) = ln(3) or x = ln(3) . When we graph f (x) = 25x and g (x) = 5x + 6, we see that they intersect at ln(5) x= ln(3) ln(5) ≈ 0.6826. x −x 6. At first, it’s unclear how to proceed with e −e = 5, besides clearing the denominator to 2 obtain ex − e−x = 10. Of course, if we rewrite e−x = e1 , we see we have another denominator x lurking in the problem: ex − e1 = 10. Clearing this denominator gives us e2x − 1 = 10ex , x and once again, we have an equation with three terms where the exponent on one term is exactly twice that of another - a ‘quadratic in disguise.’ If we let u = ex , then u2 = e2x so the equation e2x − 1 = 10ex can be viewed as u2 − 1 = 10u. Solving u2 − 10u − 1 = 0, we obtain √ √ √ by the quadratic formula u = 5 ± √ From this, we have ex = 5 ± 26. Since 5 ...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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