Mathematics
WK10 Properties of Logarithmic

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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 = ππππ (ππ )

3 = πππ2 (23 ) = 3πππ2 (8)
πππ2 (2π₯ + 1) = πππ2 (8)
2π₯ + 1 = 8
2π₯ + 1 β 1 = 8 β 1
2π₯ = 7
2π₯ 7
=
2
2
7
π₯ = πππ’π
2
π
π=
π

(b)* log x 343 = 3
ln(343)
=3
ln (π₯ )
ln(343)
ππ(π₯ ) = 3 ln(π₯ )
ln(π₯ )

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‘ππ’π
π=π

(c)6 β log(π₯ ) = 3
6 β πππ10 (π₯ ) β 6 = 3 β 6
βπππ10 (π₯ ) = β3
βπππ10 β3
=
β1
β1
πππ10 (π₯ ) = 3
Now, apply the log rule:
3 = πππ10 (103 ) = πππ10 (1000)
πππ10 (π₯ ) = πππ10 (1000)
π₯ = 1000 π‘ππ’π
π = ππππ
(d) ln ( x ) = 2
ln(π₯ ) = ln(π 2 )
π πππ£π πππ π₯ = π 2
π₯ = π 2 π‘ππ’π
π = ππ

(e) 7 log 6 (4 x) + 5 = β2
7πππ6 (4π₯ ) + 5 β 5 = β2 β 5
7πππ6 (4π₯ ) = β7
7πππ6 (4π₯) β7
=
7
7

Properties of Logarithms; Logarithmic and Exponential Equations
πππ6 (4π₯ ) = β1
apply log rule:
1
πππ6 (4π₯ ) = πππ6 ( )
6
1
4π₯ =
6
1
1
: π₯=
6
24
1
π₯=
π‘ππ’π
24
π
π=
ππ

4π₯ =

(f) log 6 36 = 5 x + 3
5π₯ + 3 = πππ6 36
5π₯ + 3 β 3 = πππ6 (36) β 3
5π₯ = β1
5π₯ β1
=
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 βlogβ key (if log base 10) or the βlnβ (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
ππ(π π₯ ) = ππ (7)
π₯ππ(π) = ln(7)
π = ππ (π)

Section 5.5 & 5.6
(b) (b)* 2e3 x = 6
π β 2π 3π₯
6
=
πβ2
πβ2
3
π 3π₯ =
π
3
ππ (π 3π₯ ) = ππ ( )
π
3
3π₯ππ(π) = ππ ( )
π
3
3π₯ = ππ ( )
π
π

π=

ππ (π)
π

(c) (c) e5 x β1 = 9
ππ 5π₯ β1...

Super_Teach12 (2461)
Boston College
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