Using the Properties of Logs
You already learned how to solve simple log equations. Now we are going a
step or two farther. These equations are solved by isolating the logarithm
(sound familiar?) and then writing an equivalent exponential equation.
Remember: Always check your answers!
Solve each of the following logarithmic equations.
1.
log 6 x  2
62  x
x  36
2.
log 2 ( x  5)  2
22  x  5
4  x5
-5
-5
x  1
Isolatethe
the
1.1.Isolate
log.(Done.)
(Done.)
log.
2. Change
2. Change
to to
exponential
exponential
form.
form.
Solve.
3.3.Solve.
4. Check.
4. Check.
3.
3 log 5 (4 x  1)  2  4
+2 +2
3____________
log 5 (4 x  1) __6
3
3
log 5 (4 x  1)  2
52  4 x  1
25  4 x  1
-1
-1
24  4__x
____
4
4
x6
1. Isolate the
log. (Done.)
2. Change to
exponential
form.
3. Solve.
4. Check.
When two or more logarithms are in the
equation, it is necessary to first combine
them to a single log using the properties of
logs!
4.
log( x)  log( x  3)  1
Product Property
log( x)( x  3)  1
101  x( x  3)
Wait! Where did that
10 come from?
x 2  3 x  10
x 2  3 x  10  0
( x  5)( x  2)  0
x  5 x  2
That’s right. It’s the
base of the common
log. It slid to the other
side.
1. Combine to a
single log.
2. Change to
exponential
form.
3. Solve.
4. Check.
5.
log 2 (3 x  23)  log 2 ( x  1)  3
3 x  23
log 2
3
x 1
Quotient Property
2. Change to
exponential
form.
3x  23
3
2 
x 1
3 x  23
8
x 1
1. Combine to a
single log.
Multiply both
sides by x+1
8( x  1)  3 x  23
3. Solve.
8 x  8  3 x  23
5 x  15
x3
4. Check.
6. Algebraically determine the intersection point of the two logarithmic functions
shown below.
y  log 3 ( x  25)  3
y  log 3 ( x  1)  5
log 3 ( x  25)  3  log 3 ( x  1)  5
log 3 ( x  25)  log 3 ( x  1)  5  3
log 3
x  25
2
x 1
x  25
9
x 1
x  25  9( x  1)
x  25  9 x  9
8 x  16
x2
Remember
how to solve
a system of
equations?
y  log 3 ( x  25)  3
y  log 3 (2  25)  3
y6
(2,6)
Are you
done?
7. Now we can find the inverses for more complicated logarithmic functions.
Given the function
f ( x)  log 5 ( x  6)  2
Find
f 1 ( x)
y  log 5 ( x  6)  2
x  log 5 ( y  6)  2
x  2  log 5 ( y  6)
5
x2
 y6
You have 2 minutes!
y  5x2  6
1
f ( x)  5
x2
6
Sometimes, there it is not necessary to write an equivalent
exponential expression in order to solve for x. (Just like when
we solve exponential equations using common bases, we can
solve log equations with a single log on each side of the
equation.
8.
log 2 x  log 2 9  log 2 45
log 2 (9 x)  log 2 45
9 x  45
x5
Product Property
1. Combine each
side to a
single log.
2. Common
Bases! No logs
needed.
3. Solve.
4. Check.
9.
log(2 x  1)  log 3  log 5
Quotient Property
2x 1
log
 log 5
3
2x 1
5
3
1. Combine each
side to a
single log.
2. Common
Bases! No logs
needed.
3. Solve.
4. Check.