Stitz-Zeager_College_Algebra_e-book

Are there other models which t the data equally well

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Unformatted text preview: x) + 3, we gather the logarithms to one side and get x+3 log2 (x + 3) − log2 (6 − x) = 3, and then use the Quotient Rule to obtain log2 6−x = 3. Rewriting this as an exponential equation gives 23 = x+3 6−x . This reduces to the linear equation 8(6 − x) = x + 3, which gives us x = 5. When we graph f (x) = we find they intersect at x = 5. ln(x+3) ln(2) and g (x) = ln(6−x) ln(2) + 3, 370 Exponential and Logarithmic Functions y = f (x) = log7 (1 − 2x) and y = g (x) = 1 − log7 (3 − x) y = f (x) = log2 (x + 3) and y = g (x) = log2 (6 − x) + 3 6. Starting with 1 + 2 log4 (x + 1) = 2 log2 (x), we gather the logs to one side to get the equation 1 = 2 log2 (x) − 2 log4 (x + 1). Before we can combine the logarithms, however, we need a common base. Since 4 is a power of 2, we use change of base to convert log4 (x+1) = log2 (x+1) = log (4) 1 2 log2 (x + 1). Hence, our original equation becomes 1 = 2 log2 (x) − 2 1 2 2 log2 (x + 1) or 1 = 2 log2 (x) − log2 (x + 1). Using the...
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