By Adams J. F.
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3 1. (a) 4 (b) -4 (c) 4 2. (a) 7 (b) -4 (c) -3 (d) 5 (d) -2 (e) 6 (f) 0 3. (a) log10 (100)13 = 13 log10 100 = 13(2) = 26 1 (b) log10 ( 100 ) = log10 1 − log10 100 = 0 − 2 = −2 (c) ln B3 = ln 3 − ln B (d) ln Ae2 = ln A = ln e2 = ln A + 2 (e) ln ABe−4 = ln A + ln B + ln e−4 = ln A + ln B − 4 (f) (log4 e)(loge 64) = log4 64 = 3 4. (a) and (c) are valid; (b) and (d) are not. 5. By definition, eln(u/v) = expressions for u v, u v. But we can also write we obtain ln uv = ln u − ln v. 57 u v = eln u eln v = e(ln u−ln v) .
A) and (b) 3. (a) An example is a modified version of the curve in Fig. 5a, with the arc AB replaced by a line segment AB. (b) A straight line. 4. Since dy/dx = b/(c + x)2 > 0, and d2 y/dx2 = −2b/(c + x)3 < 0, the curve must show b y increasing at a decreasing rate. The vertical intercept (where x = 0) is a − . when x c approaches infinity, y tends to the value a, which gives a horizontal asymptote. Thus the b range of the function is the interval [a − , a). To use it as a consumption function, we should c stipulate that: a> b [so that consumption is positive at zero income ] c b > c2 [so that M P C = dy/dx is a positive fraction throughout ] 5.
40 (d) 1 ( or 100% ) 5. When t = 0, the two functions have the same value ( the same y intercept ). Also, y1 = Aer when t = 1, but y2 = Aer when t = −1. Generally, y1 = y2 whenever the value of t in one function is the negative of the t value in the other; hence the mirror- image relationship. 3 1. (a) 4 (b) -4 (c) 4 2. (a) 7 (b) -4 (c) -3 (d) 5 (d) -2 (e) 6 (f) 0 3. (a) log10 (100)13 = 13 log10 100 = 13(2) = 26 1 (b) log10 ( 100 ) = log10 1 − log10 100 = 0 − 2 = −2 (c) ln B3 = ln 3 − ln B (d) ln Ae2 = ln A = ln e2 = ln A + 2 (e) ln ABe−4 = ln A + ln B + ln e−4 = ln A + ln B − 4 (f) (log4 e)(loge 64) = log4 64 = 3 4.
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