1001 Algebra Problems.PDF

815. d.The inputs of a logarithm must be positive.
     Since x^2 + 1 > 0 for any real number x, it fol-
     lows that its domain is the set of all real
     numbers.


816. c.The x-intercept is the point of the form
     (x,0). Since log 2 1 = 0, we conclude that the
     x-intercept is (1,0).

Set 52 (Page 121)

817. d.The range of the function g(x) = ln xis .
     Since using 2x– 1 as the input for gfor all x
     ^12 covers the same set of inputs as g, the range
     ofgis also . The graph is provided here.


818. c.The x-intercepts of the function f(x) =
     ln (x^2 – 4x+ 4) are those x-values that satisfy
     the equation x^2 – 4x+ 4 = 1, solved as follows:

x^2 – 4x+ 4 = 1

x^2 – 4x+ 3 = 0

(x– 3)(x– 1) = 0

The solutions of this equation are x= 3 and x

= 1. So, the x-intercepts are (1,0) and (3,0).

819. c.The domain of the function f(x) = ln(x^2 – 4x

+ 4) is the set ofx-values for which x^2 – 4x+ 4

0. This inequality is equivalent to (x– 2)^2 

0, which is satisfied by all real numbers xexcept

2. So, the domain is (–∞,2)∪(2,∞).

820. b.The vertical asymptote for f(x) = ln(x+ 1)

• 1 occurs at those x-values that make the input
  of the ln portion equal to zero, namely x= –1.

821. b. Using exponent and logarithm rules yields

f(g(x)) = e2(lnx) = e2ln(x ) = e2( ) lnx= elnx=

x, for all x> 0,

g(f(x)) = lne^2 x= ln(ex)^2 = ln(ex) = x,,for

all real numbers x

Hence, we conclude that these choices for f

and gare inverses.

822. b.Apply the logarithm properties and then
     solve for x:

ln(x– 2) –ln(3 – x) = 1

ln= 1

x 3 – –^2 x= e

x– 2 = e(3 – x)

x– 2 = 3e+ ex

ex+ x= 3e+ 2

(e+ 1)x= 3e+ 2

x=

Substituting this value for xin the original

equation reveals that it is indeed a solution.

^3 e+ 2

e+ 1

x– 2

3 – x

^12 ^12

–2 2 4 6

–2

–4

–6

–8

–10

10

8

6

4

2

ANSWERS & EXPLANATIONS–