# WK10 Properties of Logarithmic

*label*Mathematics

*timer*Asked: Mar 30th, 2019

*account_balance_wallet*$15

### Question Description

Attached is the document of Week 10 Guided Lecture Notes, there are problems about properties of Logarithmic. Complete the document

### Unformatted Attachment Preview

Purchase answer to see full attachment

## Tutor Answer

Hi, here is your assignment solutions :). Let me know if you need more help ;)

Properties of Logarithms; Logarithmic and Exponential Equations

From last week (5.4)

Post 1

https://mediaplayer.pearsoncmg.com/assets/bIl00_lvtNkCqnHF077MHAI55lbv7h_y

Author in Action: Solve Logarithmic Equations (6:43)

Solving Basic Logarithmic Equations

When solving simple logarithmic equations (they will get more complicated in Section 5.6) follow

these steps:

1. Isolate the logarithm if possible.

2. Change the logarithm to exponential form and use the strategies learned in Section 5.3 to solve

for the unknown variable.

Example 7*: Solve Logarithmic Equations

Solve the following logarithmic equations

(a)* log2 ( 2x +1) = 3

Apply log rule: π = ππππ (ππ )

3 = πππ2 (23 ) = 3πππ2 (8)

πππ2 (2π₯ + 1) = πππ2 (8)

2π₯ + 1 = 8

2π₯ + 1 β 1 = 8 β 1

2π₯ = 7

2π₯ 7

=

2

2

7

π₯ = πππ’π

2

π

π=

π

(b)* log x 343 = 3

ln(343)

=3

ln (π₯ )

ln(343)

ππ(π₯ ) = 3 ln(π₯ )

ln(π₯ )

Copyright Β© 2016 Pearson Education, Inc.

Section 5.5 & 5.6

ln (343) = 3 ln(π₯ )

3 ln (π₯ ) = ln (343)

3 ln(π₯ ) ln (343)

=

3

3

ln(343)

ln (π₯ ) =

3

ln(π₯ ) = ln (7)

π₯ = 7 π‘ππ’π

π=π

(c)6 β log(π₯ ) = 3

6 β πππ10 (π₯ ) β 6 = 3 β 6

βπππ10 (π₯ ) = β3

βπππ10 β3

=

β1

β1

πππ10 (π₯ ) = 3

Now, apply the log rule:

3 = πππ10 (103 ) = πππ10 (1000)

πππ10 (π₯ ) = πππ10 (1000)

π₯ = 1000 π‘ππ’π

π = ππππ

(d) ln ( x ) = 2

ln(π₯ ) = ln(π 2 )

π πππ£π πππ π₯ = π 2

π₯ = π 2 π‘ππ’π

π = ππ

(e) 7 log 6 (4 x) + 5 = β2

7πππ6 (4π₯ ) + 5 β 5 = β2 β 5

7πππ6 (4π₯ ) = β7

7πππ6 (4π₯) β7

=

7

7

Copyright Β© 2016 Pearson Education, Inc.

Properties of Logarithms; Logarithmic and Exponential Equations

πππ6 (4π₯ ) = β1

apply log rule:

1

πππ6 (4π₯ ) = πππ6 ( )

6

1

4π₯ =

6

1

1

: π₯=

6

24

1

π₯=

π‘ππ’π

24

π

π=

ππ

4π₯ =

(f) log 6 36 = 5 x + 3

5π₯ + 3 = πππ6 36

5π₯ + 3 β 3 = πππ6 (36) β 3

5π₯ = β1

5π₯ β1

=

5

5

π

π=β

π

Steps for solving exponential equations of base e or base 10

1. Isolate the exponential part

2. Change the exponent into a logarithm.

3. Use either the βlogβ key (if log base 10) or the βlnβ (if log base e) key to evaluate the variable.

Example 8*:Using Logarithms to Solve Exponential Equations

Solve each exponential equation.

(a) e x = 7

ππ(π π₯ ) = ππ (7)

π₯ππ(π) = ln(7)

π = ππ (π)

Copyright Β© 2016 Pearson Education, Inc.

Section 5.5 & 5.6

(b) (b)* 2e3 x = 6

π β 2π 3π₯

6

=

πβ2

πβ2

3

π 3π₯ =

π

3

ππ (π 3π₯ ) = ππ ( )

π

3

3π₯ππ(π) = ππ ( )

π

3

3π₯ = ππ ( )

π

π

π=

ππ (π)

π

(c) (c) e5 x β1 = 9

ππ 5π₯ β1...

*flag*Report DMCA

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors