2
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Logarithms Lesson 2
Part I – Logarithmic to Exponential Form
converting from logarithmic to exponential form:
Example 1: Convert log 2 x = y
to exponential form:
Example 2: Given 3 = log 5 x
solve for x.
Put the 3
on the other
side so you
can do the
seven rule.
7 Rule
the “seven” rule is an
easy way of remembering
the conversion. just draw a
seven as shown below, and it will
give the exponential form!
QUESTIONS:
Convert each of the following
logarithms to exponential form:
Example 3: Convert 3logb = a
to exponential form
3logb = a
a
log10 b =
3
a
3
10 = b
Whenever you
have a log
written without
a base,
it actually has
a base of 10.
1) log3a = b
2) 5 = logmn
3) log4 y = x
4) 3 = log2b
Solve for x in each of the following:
6) log3 ( 2x + 4 ) = 2
5) log2 ( x -1) = 3
Example 4: Solve for x in
log 3 ( 2x ) = y
log3 2x = y
3y = 2x
3y
x=
2
Answers:
1) 3b = a
4) b = 8
2) m5 = n
3) 4 x = y
5) x = 9
6) x = 2.5
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Logarithms Lesson 2
Part II – Exponential to Logarithmic Form
converting from exponential to logarithmic form
Example 1: Convert y = x 2 to logarithmic form
A Base is
Always a Base
First write out the logarithm with the base:
log x
=
Now fill in the rest so the “seven” rule will
give you back what you started with.
log x y = 2
remember “a base is
always a base” when
doing this conversion.
Example 2: Convert a = 10x 4 to logarithmic form
First get the x 4 by itself.
a
x4 =
10
Now set up the logarithm
log x
=
Then fill it in so the “seven” rule works.
a
log x
=4
10
Example 3: Convert y = 3x to
logarithmic form
Example 4: Convert 10 x - y =
log10
=
log10
a
=x-y
b
a
b
First simplify using exponent rules:
a=b
log b
=
log ba = m - n
now fill in the rest so
when the seven rule is
applied, you get back the
exponential function
1) y = x 3
2) 4 = 3a2
3) m5 = n
4) 2x 6 = 8
a5
a3
6) 2 x = 5
5) b =
Example 5: The logarithmic
bm
form of a = n is:
b
m-n
=
Convert each of the following from
exponential to logarithmic form
1
2
to logarithmic form:
loga
QUESTIONS:
1
Rewrite as: y 2 = 3x
log y 3x =
if you have ac = b, the a,
being the base, will also
be the base of your
logarithm!
place this first.
Answers:
4
=2
3
1) logx y = 3
2) loga
4) logx 4 = 6
5) logab = 2
3) logmn = 5
⎛5⎞ 1
6) logx ⎜ ⎟ =
⎝2⎠ 2
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Logarithms Lesson 2
Part III – Change of Base
change of base:
Change
of Base
Example 1: Evaluate log 2 3
log2 3 =
log 3
= 1.585
log 2
Example 2: Evaluate log 2
2
log
2
3 = -0.585
log2 =
3 log2
the only logs you can do in
your calculator are base 10 logs.
change of base lets you do any
logarithm in your calculator!
2
3
log b
logab =
log a
10
by writing a log as a fraction,
you automatically convert it to
base 10 logs, so now you can
type it into your calculator.
Example 3: Evaluate log5
log5 = 0.699
10
Already
base 10.
Change of
base not
needed.
Example 4: Expand log 2x (y +z)
log2x (y + z) =
log(y + z)
log2x
Example 6: Express (loga x)(log x b)
as a single logarithm
We can’t
expand log(y + z) any
further since logs are not
distributive!
⎛ log x ⎞⎛ logb ⎞ logb
(loga x)(log x b) = ⎜
= loga b
⎟⎜
⎟=
⎝ log a ⎠ ⎝ logx ⎠ loga
log(y+ z) ≠ logy+logz
Example 5: Express
as a single logarithm
log4
log7
Example 7: Evaluate the
log 8
expression: 3 2
3
log8
log2
= 33 = 27
log4
= log 4
7
log7
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Logarithms Lesson 2
Part III – Change of Base
QUESTIONS:
Evaluate using change of base :
Express each as a single logarithm
1) log4 5 =
6)
2) log5
8
9
=
7)
log5
log8
log (a - 2b)
log c
3) log 3 =
8) (logmn)(lognm)
Expand each of the following
4) log4x (y - 2z)
9) (logab)(logbc)(logc d)(logd x)
5) log(a+b) (x + y)
10) Evaluate : 5
log2 3
ANSWERS:
1) 1.16
2) - 0.073
3) 0.48
4)
5)
log(y - 2z)
log4x
6) log8 5
7) logc (a - 2b)
8) 1
log(x + y)
9) loga x
log(a + b)
10) 12.82
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Logarithms Lesson 2
Part IV – Multiplication Law
multiplication law of
logarithms
Multiplication
Law:
Example 1: Expand log ( xy )
when numbers/variables are
being multiplied inside a
logarithm, they can be expanded
by adding separate logarithms.
log ( xy ) = logx + logy
Example 2: Expand log ( 3a•2b )
log ( 3a•2b )
log a (bc) = log a b + log a c
= log(3a)+ log(2b)
= log3+ loga + log2+ logb
Example 3: Expand: log ( x + y )
log ( x + y ) = log(x + y)
remember: a log can’t be
multiplied through the brackets!
Example 4: Condense
log3+log4
log3+ log4 = log(3•4) = log12
Example 5: Condense
log(x +1)+log(x - 2)
log(x +1)+ log(x - 2)
= log(x +1)(x - 2)
= log ⎡ x 2 - x - 2⎤
⎣
⎦
Example 6: Condense
alogx +alogy
alogx + alogy
a(logx + logy)
alog(xy)
Example 7: Solve for y in the
equation: 2 = loga x + loga y
2 = loga x + loga y
2 = loga (xy)
a2 = xy
y=
a2
x
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Logarithms Lesson 2
Part IV – Multiplication Law
QUESTIONS:
Expand each of the following
Condense each of the following
1) log(abc)
4) log2 + log6
2) 2log(4x)
5) log(x + 3) + log x
3) 3log(x + y)
6) alog(xy) + a log(xz)
7) log(2x +1) + log(3x - 2)
Solve for x
8) 4 = logb x + logb y
9) 7 = logm x + logm x
2
5) log( x + 3 x )
ANSWERS:
1) log a + log b + log c
2) 2 log 4 + 2 log x
3) 3 log( x + y )
4) log12
(
6) a log x 2 yz
)
2
7) log(6 x − x − 2)
8) x =
b4
y
9) x = m7
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Logarithms Lesson 2
Part V – Division Law
division law of logarithms
⎛x⎞
⎟
⎝y⎠
Example 1: Expand log ⎜
⎛x⎞
log ⎜ ⎟ = logx - logy
⎝y⎠
Example 2: Expand log ( 3a - 2b )
log ( 3a - 2b ) = log(3a - 2b)
*division rule does
not apply here.
Division Law:
when numbers/variables are
being divided inside a
logarithm, they can be
expanded by subtracting
separate logarithms.
⎛b⎞
log a ⎜ ⎟ = log a b − log a c
⎝c⎠
x
Example 3: Expand -5log ⎛⎜ ⎞⎟
⎝3⎠
⎛x⎞
-5log ⎜ ⎟
⎝3⎠
= -5 [ logx - log3]
= -5logx +5log3
Example 4: Expand: log ( x - y )
log ( x - y ) = log(x - y)
*division rule does
not apply here.
Example 5: Condense log12 - log4
log12 - log4
⎛ 12 ⎞
= log ⎜ ⎟
⎝ 4 ⎠
= log3
Example 6: Condense log(x -1) - log(x +2)
log(x -1) - log(x + 2)
⎛ x -1 ⎞
= log ⎜
⎟
⎝ x+2 ⎠
Example 7: Condense
alogx - alogy
alogx - alogy
a(logx - logy)
alog
x
y
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Logarithms Lesson 2
Part V – Division Law
QUESTIONS:
Expand the following
1) log
Condense the following
5) log16 - log8
a
b
6) log(x + 2) - log(x - 1)
⎛a⎞
⎟
⎝2⎠
2) - 3log ⎜
7)
logx
logy
3) log ( x - y )
8) 3log27 - 3log3
4) log
(
x -2
)
2 4
2
9) log(8a b ) - log(4ab )
3 -2
-5 6
10) log(2a b ) - log(8a b )
ANSWERS:
1) log a - log b
5) log 2
2) - 3 log a + 3 log 2
6) log ⎜
3) log( x - y )
4) log( x - 2)
⎛ x+2 ⎞
⎟
⎝ x-1 ⎠
7) log y x
(Change of Base!)
8) 3 log 9
(
9) log 2ab
2
)
⎛ a8 ⎞
⎟
8⎟
⎝ 4b ⎠
10) log ⎜⎜
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Logarithms Lesson 2
Part VI – Power Law
power law of logarithms
Power Law:
Example 1: Simplify logx 2
logx 2 = 2logx
Example 2: Simplify logx 2 +logx 4
logx 2 + logx 4 = 2logx + 4logx = 6logx
Example 3: Expand (logx)2
(logx)2 = (logx)2
when there is an
exponent inside a
logarithm, it can be
taken out in front of
the logarithm.
log a b c = c log a b
Power
law does
not apply
when the
entire log is
raised to an
exponent.
Example 4: Condense 3log(xy)
3log(xy)
3
= log ( xy )
= log(x3y3 )
Example 5: Condense 2log(x -1)
2log(x -1)
= log(x -1)2
You can also write as : log(x 2 - 2x +1)
Example 6: Condense: 4loga - x
4loga - x = loga 4 - x
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Logarithms Lesson 2
Part VI – Power Law
QUESTIONS:
1) Expand : loga
3
3
2) Expand : loga + loga
3) Expand : ( loga )
7
3
4) Condense : 5log(ab)
5) Condense : 2log(a - b)
⎛a⎞
⎟-7
⎝b⎠
6) Condense : 3log ⎜
ANSWERS:
1) 3 log a
2) 10 log a
7) Simplify : ( 2log10 )
3) ( log a )
2
3
5 5
4) log a b
5) log ( a −b )
⎛a⎞
⎝b⎠
2
3
6) log ⎜ ⎟ − 7
2
8) Solve for x : log3 x = 6
7) 4
8) x = 27
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Logarithms Lesson 2
Part VII – Other Laws
other laws of logarithms
Other Laws:
Example 1: Evaluate log3x 3x
1) logax is undefined
log3x 3x =1
for x≤0.
2) loga1 = 0
Example 2: Evaluate log3x 0
3) logaa = 1
log3x 0 = Undefined
4) aloga x = x
5) logaax = x
Example 3: Evaluate log3x 1
log3x 1= 0
Example 4: Evaluate log3x (-3)
log3x (-3) = Undefined
Example 5: Evaluate log3 34
log3 34 = 4log3 3 = 4(1) = 4
Example 6: Evaluate log x -1 (x -1)2
log x-1 (x -1)2 = 2log x-1 (x -1) = 2(1) = 2
Example 10: Simplify the
Example 7: Evaluate 3log x
3
expression log a
3log3x = x
Example 8: Evaluate 4•2log 6
2
log a
( a)
4•2log2 6 = 4•6 = 24
Example 9: Simplify the expression log5 25k
= log a a 2
log5 25k
=
= log5 52k
= 2klog5 5
= 2k(1)
= 2k
x
x
⎛ 1⎞
= log a ⎜ a 2 ⎟
⎝ ⎠
= log5 (52 )k
( a)
x
x
x
log a a
2
x
= (1)
2
x
=
2
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Logarithms Lesson 2
Part VII – Other Laws
QUESTIONS:
7)
8)
9)
10)
ANSWERS:
7) 4k
3k
8)
2
9) x = 3
10) x = 4
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Logarithms Lesson 2
Part VIII – Diploma Style
diploma style logarithm questions
Example 1: Given that log3 4 = x ,
evaluate log316
Example 4: If logx =3 ,
evaluate log10x 2
log316
log10x 2
log3 4 + log3 4
log10 + logx 2
1+ 2logx
1+ 2(3)
log3 ( 4•4 )
x+x
2x
7
Example 2: If log ma =3
⎛
⎛ 1 ⎞⎞
and log m b = 4 , evaluate ⎜ log m ⎜ ⎟ ⎟
ab
⎝
⎝
⎛
⎛ 1 ⎞⎞
⎜ logm ⎜ ab ⎟ ⎟
⎝ ⎠⎠
⎝
log m (1) − log m ( ab )
log 2 A = B → A = 2B
⎛1⎞ B
log 4 A = log 4 2B = Blog 4 2 = B ⎜ ⎟ =
⎝2⎠ 2
0 − [ log m a + log m b ]
Example 6: If log a b = 0.92 ,
a
then the value of log a ⎛⎜ ⎞⎟ is:
⎝b⎠
− [ 3 + 4]
−7
y
Example 3: If x = 2 , determine
z
an expression for log x
logx
⎠⎠
Example 5: If log 2 A = B ,
then log 4 A = ?
⎛a⎞
loga ⎜ ⎟ = log a a - log a b
⎝b⎠
=1- 0.92
= 0.08
⎛y⎞
log ⎜ 2 ⎟
⎝z ⎠
logy - logz2
Example 7: If log3 x = 20 , then
⎛1 ⎞
the value of log3 ⎜ x ⎟ is:
⎝3 ⎠
logy - 2logz
⎛1 ⎞
⎛x⎞
log3 ⎜ x ⎟ = log3 ⎜ ⎟
⎝3 ⎠
⎝3⎠
= log3 x - log3 3
= 20 -1
=19
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Logarithms Lesson 2
Part VIII – Diploma Style
Example 8: If log x =3.2 and
x
log y = -0.9, then =
y
Example 11: If log b A = M ,
1
then log b 2 = ?
log10 x = 3.2 → x =10
1
= log b1- log b A2
2
A
= 0 - 2log b A
3.2
→ x =1584.89
log10 y = -0.9 → y =10 → y = 0.1259
x 1584.89
=
=12589.25
y 0.1259
-0.9
A
log b
= -2M
Example 9: If log m 9= 2 and
log8n = 2 , then log 2 (mn) = ?
Example 12: log3 (27a) = ?
logm 9 = 2 → m 2 = 9 → m = 3
= log3 27 + log3a
log8n = 2 → 8 = n → n = 64
= 3+ log3a
2
log 2 (mn) = log 2 ( 3•64 ) = log 2192 = 7.58
Example 10: If x = y 2z , then
find an expression for logz
x = y 2z → z =
x
y2
log3 (27a)
Example 13: If 10a = 4 , then
101+2a = ?
101+2a =10•10a •10a
= 10•4•4
=160
⎛ x ⎞
logz = log ⎜ 2 ⎟
⎝y ⎠
= logx - logy 2
= logx - 2logy
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Logarithms Lesson 2
Part VIII – Diploma Style
QUESTIONS:
ANSWERS:
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