MATH 180 FALL 1995

ASSIGNMENT #4

1.4, #18. We proceed as in 1.4, #17 in the last assignment. The functions are an exponential, a quadratic, and a cubic. For g(x), the ratios of consecutive values from the table are

\frac{3.74}{3.12} = \frac{4.49}{3.74} = \frac{5.39}{4.49} = \frac{6.47}{5.39} = \frac{7.76}{6.47} = 1.20

up to 2 decimal places. So g(x) is exponential with base (1.2)^{2} = 1.44. The base is 1.44 and not 1.2 since the inputs 5.0, 5.5, 6.0, 6.5, … change by .5. Indeed, if g(x) = ab^{x} and x and x + Δx are two consecutive inputs, then

\frac{g(x + Δx)}{g(x)} = \frac{ab^{x+Δx}}{ab^{x}} = b^{Δx}.

In the problem Δx = .5, b^{(.5)} = 1.2, and b = 1.44. The remaining two functions are a quadratic and a cubic power function. Since

\frac{f(9.0)}{f(8.4)} = \frac{7.29}{5.93} = 1.23, (\frac{9.0}{8.4})^{3} = 1.23,

f(x) is cubic and k(x) is necessarily quadratic.

1.6, #2, 3. The calculator or computer should show the graph of y = log 10^{x} as the line y = x, and the graph of y = 10^{log x} as the half-line y = x with x > 0. These are different graphs even though log x and 10^{x} are inverses functions of each other. The reason is that x can be any real number in log 10^{x}, whereas x is necessarily positive in 10^{log x} since the logarithm function is defined only for positive arguments.

1.6, #5. 2 = (1.02)^{t} is equivalent to t(log 1.02) = log 2 or t = 35.

1.6, #6. 7 · 3^{t} = 5 · 2^{t} is equivalent to log 7 + t(log 3) = log 5 + t(log 2). Thus

t = \frac{log 5 − log 7}{log 3 − log 2} = −.83.

1.6, #9. P = P_{0}a^{t} is equivalent to log P = log P_{0} + t(log a) or

t = \frac{log P − log P_{0}}{log a} .

Comments: The table of values of the exponential function ab^{x} in 1.4, #18 differs from the tables in 1.3, #9 and 1.4, #17. Namely, consecutive inputs differ by Δx, a number unequal to 1. The ratio of consecutive values is b^{Δx} rather than b.