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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...

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