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4/27 ALG Live Review - Solving Log Equations

4/27 ALG Live Review - Solving Log Equations

Assessment

Presentation

•

Mathematics

•

9th - 10th Grade

•

Hard

Created by

Jennifer Hedges

Used 4+ times

FREE Resource

5 Slides • 3 Questions

1

Solving Log Equations

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2

Multiple Choice

Q1:  Solve log⁡(3u+14)−log⁡5=log⁡2u\log_{ }\left(3u+14\right)-\log_{ }5=\log_{ }2u  

1

u=4

2

u=6

3

u=2

4

u=1

3

 log⁡(3u+14)−log⁡5=log⁡2u\log_{ }\left(3u+14\right)-\log_{ }5=\log_{ }2u  

  • Quotient Property:   log⁡ (3u+145)=log⁡ 2u\log\ \left(\frac{3u+14}{5}\right)=\log\ 2u  

  • One-to-One:             3u+145=2u\frac{3u+14}{5}=2u  

  • Multiply by 5:            3u+14=10u3u+14=10u  

  • Solve for u:               14=7u14=7u  

  •                                   2=u2=u  

4

Multiple Choice

Q2:  Solve

   log⁡7n=23log⁡78\log_7n=\frac{2}{3}\log_78  

1

n=1

2

n=3

3

n=6

4

n=4

5

 log⁡7n=23log⁡78\log_7n=\frac{2}{3}\log_78  

  • Power Property:     log⁡7n=log⁡7823\log_7n=\log_78^{\frac{2}{3}}  

  • Simplify:                  log⁡7n=log⁡7  382\log_7n=\log_7\ \ ^3\sqrt{8^2}  

  •                                 log⁡7n=log⁡7  364\log_7n=\log_7\ \ ^3\sqrt{64}  

  •                                 log⁡7n=log⁡74\log_7n=\log_74  

  • One-to-One:           n=4n=4  

6

Multiple Choice

Q3:   −8 + log⁡6 (x + 1) = −7-8\ +\ \log_6\ \left(x\ +\ 1\right)\ =\ -7  

1

5

2

6

3

 −10009-\frac{1000}{9}  

4

124

7

 −8 + log⁡6 (x + 1) = −7-8\ +\ \log_6\ \left(x\ +\ 1\right)\ =\ -7  

  • Isolate the log:            log⁡6 (x + 1) = 1\log_6\ \left(x\ +\ 1\right)\ =\ 1  

  • Exponential Form:      61=x+16^1=x+1  

  • Solve for x:                  5=x5=x  

8

You are ready for more practice!

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Solving Log Equations

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