Elementary Algebra
Section 4.3
Properties of Logarithms
Properties of Logarithms

Consider logb x = m and logb y = n
By definition
bm = x and bn = y
xy = (bm)(bn) = bm + n
So
logb (xy) = logb (bm + n)
=m+n
WHY?
= logb x + logb y
Product Rule for Logarithms
logb xy = logb x + logb y
for any positive real numbers b, x, y with b ≠ 1
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Properties of Logarithms
 Examples
 1. log4 (3  7) = log4 3 + log4 7
 2. log8 10 + log8 3 = log8 (10  3) = log8 30
 3. loga x2 = loga xx
= loga x + loga x
= 2 loga x
 4. 1 + log 2 + log x + log 2x2 = 1 + log 4x3
 5. ln (x – 1) + ln (x + 1) = ln ((x – 1)(x + 1))
= ln (x2 – 1)
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Properties of Logarithms

Again consider
logb x = m and logb y = n , for x, y, b positive, b ≠ 1
bm = x and bn = y
Thus
x
logb y
=
bm
logb n
b
=m–n
= logb bm –
n
WHY?
= logb x – logb y
Quotient Rule for Logarithms
x
logb y = logb x – logb y
for any positive real numbers b, x, y with b ≠ 1
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Properties of Logarithms
 Examples
 1.
log7
7
9 – log7 4
 2.
( 94 ) = log
3
log ( ) = log
16
4
4
3 – log4 16 = log4 3 – 2
 3.
Solve for x :
WHY?
log (x + 3)
=2
log (x + 1)
log (x + 3) = 2 log (x + 1)
2
WHY?
= log (x + 1)
2
x + 3 = (x + 1) = x2 + 2x+ 1
0 = x2 + x – 2 = (x + 2)(x – 1)
Solution set: { –2 , 1 }
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Properties of Logarithms

Consider
logb x = m for x, b positive, b ≠ 1
Thus
bm = x and (bm)r = xr , for any real r
and
xr = bmr = brm
So
logb xr = logb (brm)
= rm
WHY?
= r logb x
Power Rule for Logarithms
logb xr = r logb x
for any positive real numbers b, x with b ≠ 1
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Properties of Logarithms
 Examples
 1. log352 = 2 log35
 2. loga x4 = 4 loga x
 3. 3 log5(x + 1) = log5(x + 1)3
 4. x log 2 = log 2x
 5. ln 1 = loge1 = 0
 6. ln 0 = ?
Question:
Is 0 in the domain of any logarithm function ?
What does this tell you about ln 0 ?
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Properties of Logarithms

Property Recognition
Rewrite as a logarithm of a single expression :
1. log 4 + log 7
2. log 35 – log 7
3. ln 5e – ln
(
1
20e
)
(
e
20
4. log 5e – log
)
5. logb x5 – logb x3 + logb x2
logb x
6.
logb y
7. logb (x  y) in terms of logb x and logb y
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Properties of Logarithms

More Examples
1. Rewrite in expanded form: log4 (3x + 7)
Cannot be written in expanded form !
2. TRUE or FALSE : log6 36 – log6 6 = log6 30
Rewriting:
log6 (36/6) = log6 (6  5)
log6 6 = log6 6 + log6 5
0 = log6 5
Since log6 1 = 0 then log6 1 = log6 5
This implies that 1 = 5 ... a CONTRADICTION !!
Hence the given statement is FALSE !
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Properties of Logarithms

More Examples
3. TRUE or FALSE :
log3 (log2 8) =
log3 (3) =
1 =
1 =
log7 49
log8 64
log7 72
log8 82
2 log7 7
2 log8 8
2 (1)
2 (1)
1 =1
So, the given statement is TRUE !!
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Bases for Logarithms

Conversions
Can we use logb x to find loga x ?
Let
loga x = y
ay = x
logb (ay) = logb x
y logb a = logb x
(loga x)(logb a) = logb x
By definition
applying logb ...
applying power rule
replacing y
Thus
loga x =
NOTE:
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logb x
logb a
logb x
logb a
OR
loga x
logb x =
loga b
≠ logb x – logb a
Section 4.5 v5.0.1
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Bases for Logarithms

Conversion Examples
Find log3 17 on your calculator ... if you can
Having trouble ? Let’s try using a little math first ...
log 17
1.23044
log3 17 =
≈ 2.5789
=
log 3
0.477121
OR
ln 17
2.8332
log3 17 =
≈ 2.5789
=
ln 3
1.0986
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More Equations

Solve
1. Find x to the nearest whole number
e.02x = 192
ln(e.02x) = ln(192)
(.02x)ln e = ln(192)
.02x = 5.2575
x ≈ 262.9 ≈ 263
Solution set: { 263 }
2. Find x exactly
log3 (x + 1)5 = 3
3log 3 (x+1) 5 = 33
(x + 1)5 = 27
x + 1 = 271/5
x = –1 + 271/5
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Solution set: { –1 + 271/5 }
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More Equations

Solve
3. Find x exactly
log8 ( 2x + 5) + log8 3 = log8 33 = log8 (3 ∙ 11)
log8 ( 2x + 5) + log8 3 = log8 3 + log8 11
log8 ( 2x + 5) = log8 11
2x + 5 = 11
x=3
Solution set: { 3 }
4. Find x exactly
log3 2x – log3 (3x + 15) = –2
log3
(
2x
3x + 15
2x
) = –2
3x + 15
= 3–2
18x = 3x + 15
Solution set: { 1 }
x=1
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Think about it !
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Module 4
Section 4.5
Properties of Logarithms