Exponential Logarithmic and Other Common Functions

Exponential Logarithmic and Other Common Functions
Exponential Logarithmic and Other Common Functions
. f (4) = 5
4 = 625
274. f = = 3
4 = 81
275. f(–6) = = 2
6 = 64
276. (a) The domain is R and the range is (0,∞); (b) There are no zeros; (c) x =
0 is a horizontal asymptote; (d) The y-intercept is y = 1 and there are no xintercepts; (e) b = 3 > 1, therefore f is increasing on R; (f) As x approaches ∞, f
(x) approaches ∞ and as x approaches –∞, f (x) approaches 0.
277. (a) The domain is R and the range is (0,∞); (b) There are no zeros; (c) x =
0 is a horizontal asymptote; (d) The y-intercept is 1 and there are no x-intercepts;
(e) b = 0). Thus, g(x) = 6
x
is the inverse of f (x) = log6 x.
Exponential Logarithmic and Other Common Functions
(B) The logarithmic function f (x) = logb x is the inverse of the
exponential function g(x) = b
x
(b ≠ 1,b > 0). Thus, g(x) = log1.035 x is
the inverse of f (x) = (1.035)
x
.
(C) g(x) = x is the inverse of f (x) =
(D) f (x) = ln x is the natural logarithmic function that has base e; that is f
(x) = ln x = loge x, where e is the irrational constant approximately
equal to 2.718281828. Thus, its inverse is g(x) = e
x
. See the
following figure.
(E) f (x) = log x is the common logarithmic function that has base 10;
that is f (x) = log x = log10 x. Thus, its inverse is g(x) = 10
x
.
287. (a) The domain is (0,∞) and the range is (–∞,∞); (b) x = 1 is the only zero;
(c) The y-axis is a vertical asymptote; (d) The x-intercept is (1,0) and there are
no y-intercepts; (e) b = 6 > 1, so f (x) is increasing on (0,∞); (f) b = 6 > 1, so as x
approaches 0, f (x) approaches –∞ and as x approaches ∞, f (x) approaches ∞.
288. (a) The domain is (0,∞) and the range is (–∞,∞); (b) x = 1 is the only zero;
(c) The y-axis is a vertical asymptote; (d) The x-intercept is (1,0) and there are
no y-intercepts; (e) b = < 1, so f (x) is decreasing on (0,∞); (f) b = < 1, so as
x approaches 0, f (x) approaches ∞ and as x approaches ∞, f (x) approaches –∞.
289. f(e
10) = lne
10 = 10lne = 10–1 = 10
290. g(64
20) = log2 64
20 = 20log2 64 = 20(6) = 120
291. = log(100) – log(0.00000l) = log(10
2) – log(10
–6)
= 2 – (–6) = 2 6 = 8
292. g(8 · 32) = log2 (8 · 32) = log2 8 log2 32 = 3 5 = 8
293. Using the one-to-one property, u = 450.
294. (A) log8(32,768) = = 5
(B) (0.0016) = = 4
(C) log2(4,096) = = 12
(D) log1.05(2.5) = ≈ 18.78
(E) log2(400) ≈ 8.64
295.
296.
297.
298.
299.
300. Substituting A0 = 20 and k = 5,730 into the formula, A(t) = Ao
, and
evaluating at t = 5,000, omitting units for convenience, yields A(5,000) = 20
≈ 10.92 grams.
301. Evaluating the formula for x = 7.6 × 10
–4 yields pH of diet soda Z = f(7.6 ×
10
–4) = –log10(7.6 × 10
–4) ≈ 3.12.
302. To find the rate, compounded annually, that will double an investment of
$50,000 in 20 years, substitute P = $100,000, P = $50
Having a hard time figuring out how to do your assignment?
Ask our experts for help and get it done in no time!