- b.
5
4 e2–3x+ 1 < 9
4
4 e2–3x< 8
1
e2–3x< 2
ln 1 2 – 3xln 2
–2 + 0 ≤–3x–2 + ln 2
^23 ≥x> –2 +–3ln 2
2– 3 ln 2x≤^23
So, the solution set is 2– 3 ln 2,^23 .
- c.First, note that the expression ln (1 – x^2 ) is
only defined when 1 – x^2 0, which is equiva-
lent to –1x1. Since the only values of
y for which lny 0 are 0y 1, we must
determine which of these values satisfies the
more restrictive inequality 01 – x^2 1.
However, that is true for every xin the interval
(–1, 1). Hence, the solution set of the inequality
ln (1 – x^2 ) 0 is (–1, 1). - d.First, rewrite the left side as a single logarithm.
Then, convert the resulting equation to the
equivalent exponential equation and solve:
log x+ log (x+ 3) = 1
log (x(x+ 3)) = 1
(x(x+ 3)) = 10^1
x^2 + 3x– 10 = 0
(x+ 5)(x– 2) = 0
x= –5, 2However,x= – 5 cannot be a solution since – 5
is not in the domain of log x. Thus,x= 2 is the
only solution.- b.We combine the two logarithms, and then
change to the equivalent exponential equation,
which is solved as follows:
log 2 (2)x+ log 2 (x+ 1) = 2
log 2 (2x(x+ 1)) = 2
2 x(x+ 1) = 2^2
2 x^2 + 2x= 4
2 x^2 + 2x– 4 = 0
2(x– 1)(2x+ 2) = 0
x= 1, –2We must exclude x = –2 because it is not in the
domain of either logarithm in the original
equation. So,x= 1 is the only solution.- b.We combine the two logarithms, and then
change to the equivalent exponential equation,
which is solved as follows:
log (x– 2) = 2 + log (x+ 3)
log (x+ 2) – log (x+ 3) = 2
log ( ) = 2= 10^2
x+ 2 = 100(x– 3)
x=- d.Beause,bto a power is equal to zero if and
only if the power is zero, we must determine
the value ofxsuch that 3logbx= 0, or equiva-
lently logbx= 0. This equation can be written
in the exponential form x= b^0 = 1. Thus, the
solution is 1. - b.
y= e–a(b+x)
ln y= –a(b+x)
ln y= –ab– ax
ab+ ln y= – ax
- ^1 a(ab+ ln y) = x^302
99x+ 2
x– 3x+ 2
x– 3ANSWERS & EXPLANATIONS–