Stitz-Zeager_College_Algebra_e-book

51 and compare monthly compounding to continuous

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Unformatted text preview: nential. To that end, we clear denominators and get 75 1 + 3e−2t = 100. From this we get 75 + 225e−2t = 100, which leads to 225e−2t = 25, and finally, e−2t = 1 . Taking the natural log of both sides 9 360 Exponential and Logarithmic Functions gives ln e−2t = ln 1 . Since natural log is log base e, ln e−2t = −2t. We can also use 9 the Power Rule to write ln 1 = − ln(9). Putting these two steps together, we simplify 9 1 ln e−2t = ln 9 to −2t = − ln(9). We arrive at our solution, t = ln(9) which simplifies to 2 t = ln(3). (Can you explain why?) The calculator confirms the graphs of f (x) = 75 and 100 g (x) = 1+3e−2x intersect at x = ln(3) ≈ 1.099. y = f (x) = 9 · 3x and y = g (x) = 72x y = f (x) = 75 and 100 y = g (x) = 1+3e−2x x 5. We start solving 25x = 5x + 6 by rewriting 25 = 52 so that we have 52 = 5x + 6, or 52x = 5x + 6. Even though we have a common base, having two terms on the right hand side of the equation foils our plan of equating exponents or taking logs. If we stare at this long enough, we notice that we have three terms with the exponent on one term exactly twice that of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting u = 5x , we have u2 = (5x )2 = 52x so the equation 52x = 5x + 6 becomes u2 = u + 6. Solving this as u2 − u − 6 = 0 gives u = −2 or u =...
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