College Algebra, 2013a

460 Exponential and Logarithmic Functions
y = f(x) =log 117 (1 −
(
3x) and
y = f(x) =2− ln(x − 3) and
y = g(x) =log 117 x 2 − 3 ) y = g(x) =1
3. We can start solving log 6 (x+4)+log 6 (3−x) = 1 by using the Product Rule for logarithms to
rewrite the equation as log 6 [(x + 4)(3 − x)] = 1. Rewriting this as an exponential equation,
we get 6 1 =(x + 4)(3 − x). This reduces to x 2 + x − 6 = 0, which gives x = −3 andx =2.
Graphing y = f(x) = ln(x+4)
ln(6)
x = −3 andx =2.
+ ln(3−x)
ln(6)
and y = g(x) = 1, we see they intersect twice, at
y = f(x) =log 6 (x +4)+log 6 (3 − x) andy = g(x) =1
4. Taking a cue from the previous problem, we begin solving log 7 (1 − 2x) =1− log 7 (3 − x) by
first collecting the logarithms on the same side, log 7 (1 − 2x)+log 7 (3 − x) = 1, and then using
the Product Rule to get log 7 [(1 − 2x)(3 − x)] = 1. Rewriting this as an exponential equation
gives 7 1 =(1−2x)(3−x) which gives the quadratic equation 2x 2 −7x−4 = 0. Solving, we find
x = − 1 ln(1−2x)
2
and x = 4. Graphing, we find y = f(x) =
ln(7)
and y = g(x) =1− ln(3−x)
ln(7)
intersect
only at x = − 1 2 . Checking x = 4 in the original equation produces log 7(−7) = 1 − log 7 (−1),
which is a clear domain violation.
5. Starting with log 2 (x +3) = log 2 (6 − x) + 3, we gather the logarithms to one side and get
log 2 (x +3)− log 2 (6 − x) = 3. We then use the Quotient Rule and convert to an exponential
equation
( ) x +3
log 2 =3 ⇐⇒ 2 3 = x +3
6 − x
6 − x
This reduces to the linear equation 8(6 − x) =x + 3, which gives us x = 5. When we graph
f(x) = ln(x+3)
ln(2)
and g(x) = ln(6−x)
ln(2)
+ 3, we find they intersect at x =5.