4.
Chapter 12, Math 095
Disclaimer: This file is a compilation of problems created by Math 095 instructors at WWCC. These problems are meant only to give students extra practice in preparation for exams or for understanding a particular
topic. We do not promise that the problems below are problems that will appear on your exam in Math 095.
1.
Given f HxL =
3
è!!!!!!!!
!!
x + 2 , find f -1 HxL.
Simplify each expression.
HaL
HbL
3
è!!!!!!!!
!!
x+2
3
è!!!!!!!!
!!
y= x+2
è!!!!!!!!
3
!!
x= y+2
è!!!!!!!!!!
3
! 3
3
x =I y+2M
f HxL =
log2 32
log2 32 = log2 25
= 5 log2 2
= 5ÿ1
log16 ÅÅÅÅ18
log16 ÅÅÅÅ18 = x
ÅÅÅÅ18
1
ÅÅÅÅ
ÅÅ
23
2-3
f -1 HxL = x3 - 2
log16 ÅÅÅÅ18 = - ÅÅÅÅ34
5.
52 x
Solve for x. log3 x = 2
log3 x = 2
= 125
32 = x
9=x
52 x = 53
2x=3
x = ÅÅÅÅ32
3.
Sketch the graph of the function. f HxL = 2 + 4
6.
Solve for x. log x 81 = 4
log x 81 = 4
x
x
y
-2
1
17
2-2 + 4 = ÅÅÅ
Å + 4 = ÅÅÅÅÅ
Å
4
4
-1
1
9
2-1 + 4 = ÅÅÅ
Å + 4 = ÅÅÅ
Å
2
2
0
20 + 4 = 1 + 4 = 5
1
21
2
22 + 4 = 4 + 4 = 8
x
24 x
- ÅÅÅÅ34 = x
y = x3 - 2
Solve for x. 52 x = 125
= H24 L
=
-3 = 4 x
x3 = y + 2
x3 - 2 = y
2.
= 16x
x4 = 81
4
è!!!!!
4
è!!!!!
!
x4 = 81
x=!3
We don't have logarithms with negative bases so the solution is: x = 3
7.
+4 = 2+4 = 6
Write the expression as a single logarithm. 2 log3 x - 4 log3 Hx + 1L + log3 Hx - 2L
2 log3 x - 4 log3 Hx + 1L + log3 Hx - 2L = log3 x2 - log3 Hx + 1L4 + log3 Hx - 2L
x
= log3 ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ + log3 Hx - 2L
Hx+1L4
2
x
= log3 A ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ ÿ Hx - 2LE
Hx+1L4
8
2
6
-3
-2
-1
8.
Solve for x. 63 x = 5
4
63 x = 5
2
ln 63 x = ln 5
3 x ln 6 = ln 5
1
2
3
ln 5
x = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
3 ln 6
x º 0.299415
Check:
log2
è!!!
x= 8:
è!!!
è!!!
8 + log2 I2 8 M - 3 = 1 ?
log2 H23ê2 L + log2 H25ê2 L - 3 = 1 ?
9.
ÅÅÅÅ32 + ÅÅÅÅ52 - 3 = 1 ?
è!!!
è!!!
x = - 8 : log2 I- 8 M is undefined
è!!!
Solution : x = 8
Solve for x. 3 ÿ 4 x+5 - 1 = 1
3 ÿ 4 x+5 - 1 = 1
3 ÿ 4 x+5 = 2
4 - 3 = 1 ? Yes
12. Find the value of the following logarithmic expressions.
HaL
4 x+5 = ÅÅÅÅ23
log5 H100L - log5 20
ln 4 x+5 = ln ÅÅÅÅ23
Hx + 5L ln 4 = ln ÅÅÅÅ23
100
log5 H100L - log5 20 = log5 H ÅÅÅÅ
ÅÅÅÅÅ L
20
= log5 5
x ln 4 + 5 ln 4 = ln ÅÅÅÅ23
HbL
x ln 4 = ln ÅÅÅÅ23 - 5 ln 4
=1
log3 27
ln ÅÅ2ÅÅ -5 ln 4
log3 27 = log3 33
=3
3
x = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
ln 4
10.
HcL
Solve for x. logH5 xL - logHx + 1L = 4
logH5 xL - logHx + 1L = 4
5x
logH ÅÅÅÅ
ÅÅÅÅÅÅ L
x+1
5x
ÅÅÅÅ
ÅÅÅÅÅ
x+1
5x
ÅÅÅÅ
ÅÅÅÅÅ
x+1
=4
1
log2 ÅÅÅÅ
= log2 2-3
8
4
= 10
HdL
= 10000
5 x = 10000 Hx + 1L
5 x = 10000 x + 10000
-9995 x = 10000
10000
x = - ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ
9995
10000
Check: logH5 H- ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ LL is undefined. This means there is no solution
9995
11.
log2 ÅÅÅÅ18
Solve for x. log2 x + log2 H2 xL - 3 = 1
log2 x + log2 H2 xL - 3 = 1
log2 x + log2 H2 xL = 4
log2 Hx ÿ 2 xL = 4
ln
13. Write 641ê3 = 4 in logarithmic form.
641ê3 = 4
log64 4 = ÅÅÅÅ13
x
14. Find f -1 HxL if f HxL = ÅÅÅÅ
- 5. Use f -1 HxL notation.
2
x
f HxL = ÅÅÅÅ
-5
2
y = ÅÅÅÅ2x - 5
x = ÅÅÅÅ2y - 5
2 x2 = 2 4
x + 5 = ÅÅÅÅ2y
2 Hx + 5L = y
y = 2 Hx + 5L
2 x2 = 16
è!!!
x=! 8
x2 = 8
è!!!
x= 8:
è!!!
è!!!
log2 8 + log2 I2 8 M - 3 = 1 ?
log2 H23ê2 L + log2 H25ê2 L - 3 = 1 ?
è!!!
è!!!
x = - 8 : log2 I- 8 M is undefined
è!!!
Solution : x = 8
ÅÅÅÅ32 + ÅÅÅÅ52 - 3 = 1 ?
4 - 3 = 1 ? Yes
3
è!!!
‰ = ln ‰1ê3
= ÅÅÅÅ13
log2 H2 x2 L = 4
Check:
= -3
3
è!!!
ln ‰
f -1 HxL = 2 x + 10
15.
Solve for x. Give the exact value and approximate correct to four decimal places.
HaL
53 x = 5x+1
53 x = 5 x+1
3x= x+1
2x=1
x = ÅÅÅÅ12 = 0.5
HbL
log5 Hx + 1L = 3
HbL
log5 Hx + 1L = 3
4
53 = x + 1
125 = x + 1
124 = x
HcL
gHxL = log2 x
2
-4
-2
2
6x = 2
6x = 2
ln 6 x = ln 2
x ln 6 = ln 2
ln 2
x = ÅÅÅÅÅÅÅÅÅ º 0.3869
ln 6
HdL
-4
17. Given the of f HxL below, plot f -1 HxL on the same axes.
4
3 log2 x = 6
2
3 log2 x = 6
log2 x = 2
-4
-2
22 = x
x=4
HeL
4
-2
log6 x + log6 Hx - 5L = 1
4
-4
log6 x + log6 Hx - 5L = 1
log6 HxHx - 5LL = 1
log6
2
-2
Hx2
4
- 5 xL = 1
2
61 = x 2 - 5 x
0 = x2 - 5 x - 6
0 = Hx - 6L Hx + 1L
x = 6 or x = -1 Hno negative input in logL
-4
-2
2
4
-2
x=6
-4
18.
16. Graph the following functions.
HaL
f HxL = 2
x
$4000 is invested in an account earning 3% interest compounded monthly.
HaL
2
-2
2
-2
-4
A = PH1 + ÅÅÅÅnr Ln t
0.03
A = 4000 H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
4
-4
Find the amount in the account after 6 years, correct to the nearest penny.
4
HbL
12 H6L
A = $4787 .79
How long will it take for the account to reach $5000?
A = PH1 + ÅÅÅÅnr Ln t
0.03
5000 = 4000 H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
5000
0.03
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ = H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
4000
12
0.03
ÅÅÅÅ54 = H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
lnH ÅÅÅÅ54 L
lnH ÅÅÅÅ54 L
=
=
t=
19.
x+5
Find f -1 HxL, f HxL = ÅÅÅÅ
ÅÅÅÅÅÅ
x-3
4
2
12 t
12 t
-4
0.03 12 t
lnH1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
0.03
12 t lnH1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
lnH5ê4L
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ º 7.45 years
12 lnH1+0.03ê12L
so it will take 7 years, 6 months
22.
Solve: log2 H3 x - 2L - 5 = 3
f
HxL =
1
9 x = ÅÅÅÅ
ÅÅ
27
1
9x = ÅÅÅÅ
ÅÅ
27
1
= ÅÅÅÅ
ÅÅ
33
3 = 3-3
2 x = -3
2x
HbL
x = - ÅÅÅÅ32
log3 x = 4
log3 x = 4
x = 34
x = 81
23. Solve. Write the exact solution and an approximation to the solution accurate to four decimal places.
33 x-5 = 7
3 x - 2 = 28
3 x - 2 = 256
3 x = 258
33 x-5 = 7
ln 33 x-5 = ln 7
H3 x - 5L ln 3 = ln 7
3 x ln 3 - 5 ln 3 = ln 7
3 x ln 3 = ln 7 + 5 ln 3
258
x = ÅÅÅÅ
ÅÅÅÅÅ
3
x = 86
21.
g.
4
x
H32 L
3 x+5
ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x-1
log2 H3 x - 2L - 5 = 3
log2 H3 x - 2L = 8
2
Find x. Be sure to show your work.
HaL
y+5
ÅÅÅÅ
ÅÅÅÅÅ
y-3
3 x+5
y = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
x-1
20.
-2
-4
xHy - 3L = y + 5
xy-3x = y+5
xy-y=3x+5
yHx - 1L = 3 x + 5
-1
gHxL
-2
x+5
f HxL = ÅÅÅÅ
ÅÅÅÅÅÅ
x-3
x=
fHxL
12 t
Graph: f HxL = 3 x and gHxL = log3 x below. Be sure to make it clear which graph is f and which graph is
ln 7+5 ln 3
x = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ
3 ln 3
x = 2.25708
24.
Write as one logarithm and then simplify: 2 log16 6 - 2 H2 log16 2 + log16 3L
2 log16 6 - 2 H2 log16 2 + log16 3L = 2 log16 6 - 2 Hlog16 22 + log16 3L
= 2 log16 6 - 2 log16 H22 ÿ 3L
= 2 log16 6 - log16 H22 ÿ 3L
2
= 2 log16 6 - log16 144
= log16 62 - log16 144
36
= log16 ÅÅÅÅ
ÅÅÅÅÅ
144
= log16 ÅÅÅÅ14
= log16 16-1ê2
= -1 ê 2
25.
Write
as
a
sum,
difference,
x +3 x-10
log8 J ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ N
3
è!!!!!!!!!
and/or
multiple
of
logarithms
2
x-1
x +3 x-10
log8 J ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅ N = log8 Hx2 + 3 x - 10L - log6
3
è!!!!!!!!!
2
x-1
3
è!!!!!!!!
!!
x-1
= log8 @Hx - 2L Hx + 5LD - ÅÅÅÅ13 log6 Hx - 1L
1
= log8 Hx - 2L + log8 Hx + 5L - ÅÅÅÅ
log6 Hx - 1L
3
26.
Solve. log3 Hx + 2L + log3 Hx + 3L - log3 Hx - 6L = 1
log3 Hx + 2L + log3 Hx + 3L - log3 Hx - 6L = 1
log3 @Hx + 2L Hx + 3LD - log3 Hx - 6L = 1
Hx+2L Hx+3L
log3 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ = 1
x-6
Hx+2L Hx+3L
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ = 31
x-6
Hx + 2L Hx + 3L = 3 Hx - 6L
x2 + 5 x + 6 = 3 x - 18
x2 + 2 x + 24 = 0
è!!!!!!!!!!!!!!!!
!!!!!!!!!!!
2
-2! 2 -4 H1L H24L
x = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅ
2 H1L
è!!!!!!!!!
-2! -92
x = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
2
No Solution HimaginaryL
27.
Given logb 2 = 0.431, logb 3 = 0.683, and logb 11 = 1.490, find:
HaL
HbL
logb 396
logb 396 = logb H22 ÿ 32 ÿ 11L
= logb 22 + logb 32 + logb 11
= 2 logb 2 + 2 logb 3 + logb 11
= 2 H0.431L + 2 H0.683L + H1.490L
= 3.718
18
logb ÅÅÅÅ
ÅÅ
11
18
logb ÅÅÅÅ
ÅÅ = logb 18 - logb 11
11
= logb 2 ÿ 32 - logb 11
= logb 2 + logb 32 - logb 11
= logb 2 + 2 logb 3 - logb 11
= H0.431L + 2 H0.683L - H1.490L
= 0.307