Set 51 (Page 119)
- b.Using the logarithm rules yields
3ln(xy^2 ) = – 4ln(x^2 y) + ln(xy) =
ln(xy^2 )^3 = – ln(x^2 y)^4 + ln(xy) =
ln+ ln(xy) =
ln =
ln =
ln
- b.log 8 2 + log 8 4 = log 8 (2.4) = log 8 8 =1
- c.4 log 9 3 = log 9 (3^4 ) = log 9 81 = 2
- b.ln18x^3 – ln6x= ln(^168 xx
^3
) = ln3x^2 - b.log 7 429 – log 7 ^27 = log 7 ( 429 72 ) = log 7 ( 429 ^27 )
= log 7 ^17 = –1 - d.3 log 4 ^23 + log 4 27 = log 4 (^23 )^3 + log 4 27 = log 4
287 + log 4 27 = log 4 ( 287 27) = log 4 8 = ^32 - c.log (2x^3 ) = log 2 + logx^3 = log 2 + 3 logx
- d.
log 2 (^8 xy 2 z
^4
) = log 2 (8yz^4 ) – log 2 x^2 = log 2 8 + log 2
y+ log 2 z^4 – log 2 x^2 = 3 + log 2 y+ 4log 2 z– 2log 2 x - d.
^32 log 2 4 – ^23 log 2 8 + log 2 2 =
^32 log 2 22 – ^23 log 2 23 + log 2 2 =
^32 (2) – ^23 (3) + 1 =
3 – 2 + 1 =
2 - d.
3 logb(x+ 3)–1– 2logbx+ logb(x+ 3)^3 =
–3 logb(x+ 3) – 2logbx+ 3logb(x+ 3) =
- 2logbx=
logbx–2=
logb()
811. c.
ln (2x+ 1)(x^2 + 3)^4 =
ln (2x+ 1) + ln(x^2 + 3)^4 =
ln (2x+ 1) + 4ln(x^2 + 3) =
ln 2 + ln (x+ 1) + 4ln (x^2 + 3) =
ln 2 + ln (x+ 1) + 4ln (x^2 + 3) =
ln 2 + ^12 ln (x+ 1) + 4ln (x^2 + 3)
- c.
log 3
= log 3 (x^2 2 x– 1) – log 3 (2x+ 1)
= log 3 (x^2 ) + log 3 ( 2 x– 1) – log 3 (2x+ 1)
= log 3 (x^2 ) + log 3 (2x– 1) – log 3 (2x+ 1)
= 2log 3 (x) + ^12 log 3 (2x– 1) – ^32 log 3 (2x+ 1)
- b.As the x-values decrease toward zero, the
y-values on the graph off(x) = ln xplunge
downward very sharply, as can be seen in the
following graph: - b.The inputs for logarithmic functions must
be positive. Since the input in this case is –x,
then the inequality –x0 must be satisfied.
Clearly –x0 =>x< 0. Thus the domain ofk
is (–∞, 0).
1 2 3 x
–3–2–1 4 5 6 7 8 910
–2
–4
4
2
1
–1
–3
–5
3
^12 ^32
^32
^32
x^2 ^2 x– 1
(2x+ 1)^32
^12
^1
x^2
y^3
x^4
x^3 y^6 xy
x^8 y^4
(xy^2 )^3 (xy)
(x^2 y)^4
(xy^2 )^3
(x^2 y)^4
ANSWERS & EXPLANATIONS–