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TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ......

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TEST 6 Review 1) 2 5+3x = 1/16 2 5+3x = 0.0625 log 2 [2 5+3x ] = log 2 [0.0625] Take base-2 log on both sides. [5+3x]log 2 (2) = log 10 (0.0625)/log 10 (2) Use Power Rule and change-of-base formula 5+3x = log 10 (0.0625)/log 10 (2) Note: log 2 (2) = 1 5+3x = -4 5+3x - 5 = -4 5 3x = -9 x = -3
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Page 1: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

TEST 6 Review

1)

25+3x

= 1/16

25+3x

= 0.0625

log2[25+3x

] = log2[0.0625] Take base-2 log on both sides.

[5+3x]log2(2) = log10(0.0625)/log10(2) Use Power Rule and change-of-base formula

5+3x = log10(0.0625)/log10(2) Note: log2(2) = 1

5+3x = -4

5+3x - 5 = -4 – 5

3x = -9

x = -3

Page 2: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

2) y = 4(4x-2)

Page 3: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

3) y = 2x

Page 4: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

4) (4/5)x = 16/25

0.8x = 0.64

log0.8[0.8x] = log0.8[0.64] Take base-0.8 log of both sides.

xlog0.8(0.8) = log10(0.64)/log10(0.8) Use Power Rule and change-of-base formula

x = log10(0.64)/log10(0.8) Note: log0.8(0.8) = 1

x = 2

5) 41+2x

= 64

log4[41+2x

] = log4[64] Take base-4 log of both sides.

[1+2x]log4(4) = log10(64)/log10(4) Use Power Rule and change-of-base formula

1+2x = log10(64)/log10(4) Note: log4(4) = 1

1+2x = 3

1+2x - 1= 3 – 1

2x = 2

X = 1

6) f(x) = ax

If a is allowed to be negative, the function could not be defined for

all values of x, such as negative values.)

Page 5: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

7) y = (1/2)x

Page 6: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

8) 4x = 1/16

4x = 0.015625

log4[4x] = log4[0.015625] Take base-4 log of both sides.

xlog4(4) = log10(0.015625)/log10(4) Use Power Rule and change-of-base formula

x = log10(0.015625)/log10(4) Note: log4(4) = 1

x = -3

3

3 1 1/33

3

1/3

27

9) Write 27 3 in log form

Note: 27 27 27

27 3

27 3

log 3 1/3

10) Log4(7)

Page 7: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

When t = 0, f(t) = 8

12) log5(1/25) = -2

5-2 = 1/25 13) 62 = 36 log6(36) = 2

Page 8: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

14) Evaluate log8(1/512)

= log10(1/512) / log10(8) = -3

15)

10

5

10

1log

1 25Evaluate log 2

25 log 5

16) Write 43 = 64 in log form log4(64) = 3 17) Write logt(t) - logt(s) + 3 logt(u) as a single logarithm. logt(t) - logt(s) + logt(u)3

3

t t t

3

t t

3

t

log (t) + log (u) - log (s)

log (t u ) - log (s)

t ulog

s

Page 9: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

8 9

5 2

1/8 1/9

5 2

1/8 1/9 2

5 5 5

5 5 5

18) Write log as sum and/or difference of logarithms.

log

log log log

1 1log log 2log

8 9

x y

z

x y

z

x y z

x y z

2

b

2

log 49

b

log

log 49

19) Evaluate 2

log property: log

log property: b

Thus, 2 49

x

x

b x

x

Page 10: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

2

6n 3

12

6n 3

1/ 62

n 3

2

n 3

2 3

n n n

n n n

n n n

n n

920) Write log as sum and/or difference of logarithms.

9log

9log

1 9log

6

1log 9 log - log

6

1log 9 2log - 3log

6

1 2 3log 9 log log

6 6 6

1 1log 9 log

6 3

x

z

x

z

x

z

x

z

x z

x z

x z

n

1log

2x z

21)Write log log as a single logarithm.

log

m m

m

m n

m n

Page 11: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

12

1/ 2

12

1/ 2

12 12 12

112 12 122

1722)Write log as sum and/or difference of logarithms.

17log

log 17 log log

log 17 log log

x

y

x

y

x y

x y

a a a a

3/5 1/ 2 6

a a a a

1/ 2 3/5 6

a a a a

1/ 2 3/5 6

a a a a

1/ 2 3/5 6

a a

1/ 2

a 3/5 6

3 123) Write log log log 6log as a single logarithm.

5 2

log log log log

log log log log

log log log +log

log log

log

x y w z

x y w z

x w y z

x w y z

x w y z

x w

y z

Page 12: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

2 5

4

2 5

4 4 4

4 4 4

24)Write log as sum and/or difference of logarithms.7

log log log 7

2log 5log log 7

x y

x y

x y

25) ln 0.986 = -0.0140989243795

10

26) log(0.00314)

log (0.00314) 2.503070351926

Page 13: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

6 6

2

6 6

27) Solve log ( 1) log ( 4) 2

Note: 2 written in terms of log base-6 is

2 = log (6 ) log (36)

x x

6 6

6 6 6

6 6

2

2

2

2

2

log ( 1) log ( 4) 2

log ( 1) log ( 4) log (36)

log ( 1) ( 4) log (36)

( 1) ( 4) 36

4 4 36

3 4 36

3 4 36 36 36

3 40 0

3 40 0

( 8)( 5) 0

set x + 8 = 0 set x -

x x

x x

x x

x x

x x x

x x

x x

x x

x x

x x

5 = 0

x = -8 x = 5

Page 14: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

6 6

6 6

6 6

6 6

6

Check answers:

For x = -8

log ( 1) log ( 4) 2

log ( 8 1) log ( 8 4) 2

log ( 9) log ( 4) 2

Since log of a negative number is undefined,

-8 is an extraneous solution.

For x = 5

log ( 1) log ( 4) 2

log (5 1)

x x

x x

6

6 6

6

6

log (5 4) 2

log (4) log (9) 2

log (4 9) 2

log (36) 2

2 2

Solution set is {5}

Page 15: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

5 5

3

5 5

5 5 5

5 5 5

5 5

28) Solve log ( 1) 3 log (6 4)

Note: 3 written in terms of log base-5 is

3 = log (5 ) log (125)

log ( 1) log (125) log (6 4)

log ( 1) log (125) log (6 4)

1log log (6 4)

125

1

1

x x

x x

x x

xx

x

6 425

1(125) 6 4 125

125

1 750 500

1 750 500

0 750 500

0 749 501

0 501 749 501 501

501 749

501 749

749 749

501

7

1 1

49

1

x

xx

x x

x x

x

x

x

x

x

x

x

x

x

Page 16: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

5 5

5 5

5 5

Check Answer:

501

749

log ( 1) 3 log (6 4)

501 501log 1 3 log 6 4

749 749

501log negative number 3 log 6 4

749

501 is an extraneous solution.

749

x

x x

x

229) ln(4.37 10 ) ln(0.0437) 3.130407176

1030) log 2.18 log 2.18 0.3384564936

Page 17: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

31) 193-x = 26

log19[193-x

] = log19[26] Take base-19 log of both sides.

[3-x]log19(19) = log10(26)/log10(19) Use Power Rule and change-of-base formula

3-x = log10(26)/log10(19) Note: log19(19) = 1

3-x = 1.10652540639298

3-x-3= 1.10652540639298 -3

-x = -1.89347459360702

X = 1.89347459360702

32) 6x+1 = 29

log6[6x+1

] = log6[29] Take base-6 log of both sides.

[x+1]log6(6) = log10(29)/log10(6) Use Power Rule and change-of-base formula

x+1 = log10(29)/log10(6) Note: log6(6) = 1

x+1 = 1.87932358545715

x+1-1 = 1.87932358545715 -1

x = 0.87932358545715;

33) 20x = 54

log20[20x] = log20[54] Take base-20 log of both sides.

xlog20(20) = log10(54)/log10(20) Use Power Rule and change-of-base formula

x = log10(54)/log10(20) Note: log20(20) = 1

x = 1.33155558718601

Page 18: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

34)

log₄(x²) = log₄(5x + 14)

(x²) = (5x+14)

x² = 5x + 14

x² - (5x + 14) = 5x + 14 - (5x + 14)

x²-5x - 14 = 0

(x-7)(x+2) = 0

Set x-7 = 0 set x+2 = 0

X = 7 x = -2

Check answers:

log₄(x²) = log₄(5x + 14)

log₄((7)²) = log₄(5(7) + 14)

log₄(49) = log₄(49)

log₄((-2)²) = log₄(5(-2) + 14)

log₄(4) = log₄(4)

Solution set is {7, -2}

Page 19: TEST 6 Review 1) - simulation-math.com fileTEST 6 Review 1) 25+3x = 1/16 25+3x = 0.0625 log 2 [2 ... X = 1 6) f(x) = ax ... log 9 2log - 3log 6 1 2 3 log 9 log log

35)

log3(x) = 5

Note: 5 can be written as 5 = log3(35) = log3(243)

Rewrite log3(x) = 5 as follows:

log3(x) = log3(243)

x = 243

Solution set is {243}

Check answer:

log3(x) = 5

log3(243) = 5

5=5


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