L5 – 7.4 – Solving Logarithmic Equations
MHF4U
Jensen
Part 1: Try and Solve a Logarithmic Equation
Solve the equation log $ + 5 = 2 log($ − 1)
Hint: apply the power law of logarithms to the right side of the equation
logle
5)
+
=
2logle
↑
g(e 5) oglee-11
x
+
0
0
x
5
=
=
=
=
does
"
for
=
+
x+ 5
1)
-
( 1)
+
st
x2
x"
-
-
x
30
+
-
1
For
4
5
-
x
,
=
-
I
this means
?
solutions
0
4413
-
5
1
-
>
solutions
0
4 only
!
o
=
Note:
If log . / = log . 0, then / = 0.
:
xC >
1
>
5
-
log(oc-1)
x
b u(x + 1)
sides
:
+
x7
2x
both
>
-
=
4
loglec
for
work
this
-1
is
Part 1: Solve Simple Logarithmic Equations
an
3I
is
extraneous solution
To complete this lesson, you will need
to remember how to change from
logarithmic to exponential:
Example 2: Solve each of the following equations
1 = log 2 $ à
a) log($ + 4) = 1
Method 1: re-write in exponential
form
x+ 4
=
4
=
x +
La
=
Method 2: express both sides as a
logarithm of the same base
log(x
g(x
10
10
6
x+
x
+
x)
6)
-
H
-
>
x=
0
4
/
Sol'n is
not
extraneous
4
+
4)
+
4)
=
6
10
=
=
1
g10
10
b) log 3 (2$ − 3) = 2
Part 2: Apply Factoring Strategies to Solve Equations
Example 3: Solve each equation and reject any extraneous roots
a) log $ − 1 − 1 = − log $ + 2
1es0
6
>
b) log √$ 6 + 48$ = ?
loglec"
+
48ed)" =
Elogle"X318)
+
loglec
48xd
+
ec+ 48x
=
=
() x3
2
log()
x
x+ 48x
ech +
100
=
H8x -100
=
(x + 50)(x 2)
-
=
-
plug
50
,
x
=
9(x
9x
=
=
-
-
xeu
u)
36
Gx-x
2
=
ea
* needs
to
than 4 d in
2
soling into
2
=
&
Potential Soln
x
=
36
0
=
x
x
10
=
it
greater
this
is
equation
Example 4: If log 5 0 = 3, then use log rules to find the value of…
a) log 5 /0 6
loga (a) logalb)
loga(a) 2logab
+
+
1
41
L
2(3)
+
+
=
6
7
b) log 2 /
loga 9
logale
I
3
n
c) log ? $ − log ? ($ − 4) = 2
Hint: need to change the base
789 : ; =
g
case
,
e > u
ignore