Download - 5.6 Laws of Logarithms 5.7 Exponential Equations; Changing Base Objectives: 1.Compare & recall the properties of exponents 2.Deduce the properties of logarithms.

5.6 Laws of Logarithms5.7 Exponential Equations; Changing Base

Objectives:1. Compare & recall the properties of

exponents

2. Deduce the properties of logarithms from/by comparing the properties of exponents

3. Use the properties of logarithms

4. Solve the exponential equations

Since the logarithmic function y = logb x is the inverse of the exponential function y = bx, the laws of logarithms are very closely related to the laws of exponent.

Pre-Knowledge

For any b, c, u, v +, and b ≠ 1, c ≠ 1, there exists some x, y , such that

u = bx, v = by

By the previous section knowledge, as long as taking

x = logbu, y = logbv

1. Product of Power

am an = am+n

1. Product Property

logbuv = logbu + logbv

Proof

logbuv = logb(bxby)= logbb x+y = x + y

= logbu + logbv

Pre-Knowledge


u = bx, v = by



2. Quotient Property

2. Quotient of Power

a

aa

m

nm n

vuv

ubbb logloglog

Proof

vuyxbb

b

v

ubb

yx by

x

bb logloglogloglog

Pre-Knowledge


u = bx, v = by



3. Equal Power

am = an iff m = n

3. Equal Property

logbu = logbv iff u = v

Proof

logbu – logbv = 0

0log v

ub 10 b

v

uvu

4. Power of Power

(am)n = amn

4. Power Property

logbuk = k logbu

Proof

logbuk = logb(bx)k = logbb kx = kx = k logbu

Pre-Knowledge


u = bx, v = by



5. Change-of-Base Formula

b

u u

c

cb log

loglog

Proof Note that

bx = u, logbu = x

Taking the logarithm with base c at both sides:

logcbx = logcu or x logcb = logcu

b

u x

c

c

log

log

b

u u

c

cb log

loglog

6. Reciprocal Formula

Proof

bb

u u

uu

ub log

1

log

loglog

b u

ub log

1log

7. Raise Power Formula

Proof

k

bk

k

b ub

u

bk

uk

b

uu klog

ln

ln

ln

ln

ln

lnlog

0 , ,loglog kku u k

bb k

Example 1 Assume that log95 = a, log911 = b, evaluate

a) log9 (5/11)

b) log955

c) log9125

d) log9(121/45)

e) log9275

Example 2 Expanding the expression

a) ln(3y4/x3)

ln(3y4/x3) = ln(3y4) – lnx3 = ln3 + lny4 – lnx3

= ln3 + 4 ln|y| – 3 lnx

b) log3125/6x9

log3125/6x9 = log3125/6 + log3x9

= 5/6 log312 + 9 log3x

= 5/6 log3(3· 22) + 9 log3x

= 5/6 (log33 + log322) + 9 log3x

= 5/6 ( 1 + 2 log32) + 9 log3x

Example 3 Condensing the expression

a) 3 ( ln3 – lnx ) + ( lnx – ln9 )

3 ( ln3 – lnx ) + ( lnx – ln9 )

= 3 ln3 – 3 lnx + lnx – 2 ln3

= ln3 – 2 lnx

= ln(3/x2)

b) 2 log37 – 5 log3 x + 6 log9 y2

2 log37 – 5 log3 x + 6 log9 y2

= log349 – log3 x5 + 6 ( log3 y2/ log39)

= log3(49/x5) + 3 log3 y2

= log3(49y6/x5)

Practice

A) P. 199 Q 7 – 18

B) P. 199 Q 19 – 20 How do you change to make it to be true?

True or False log a y = – log 1/a y

c) P. 200 Q 7 – 27 (odd)

Example 4 Calculate log48 and log615 using common and natural logarithms.

a) log48

log48 = log8 / log4 = 3 log2 / (2 log2)

= 3/2

log48 = ln8 / ln4 = 3 ln2 / (2 ln2) = 3/2

b) log615 = log15 / log6 = 1.511

Example 5. Express 3

logb

M

Nin terms of log logb bM and N

31log log

2 b bM N

13log log

2 b bM N

13 3 2

log logb b

M M

N N

31log

2 b

M

N

More on Expand/Condense logarithmic expressions

Example 6 Expand

1444

6 6 6 64

6 6 6

26 6 6 6

6 6 6

72log log 72 log log ( )

1log (36 2) log ( ) 4 log | |

41

log (6 ) log (2) log ( ) 4 log | |4

12 log (2) log ( ) 4 log | |

4

xx y

y

x y

x y

x y

4

4

6

72log

y

x

Example 7 Expand in terms of sums and differences of logarithms

3 43 4 2

2log log ( ) loga a a

w yw y z

z

3 4

2loga

w y

z

3 4 2log log loga a aw y z


zyw aaa log2log4log3

222

2

3log log 3 log 1

1a a a

x xx x x

x

22log log 3 log 1a a ax x x

Example 8 Expand to express all powers as factors

2

2

)1(

3log

x

xxa


12log | | log ( 3) 2log 1

2a a ax x x

Example 9 Condense to a single logarithm. 1

6log 2log log3b b bx y z

6 2 1/ 316log 2log log log log log

3b b b b b bx y z x y z

61/ 3

2log logb b

xz

y

6 1/ 3 6 3

2 2log , or logb b

x z x z

y y


Assignment:5.6

P. 196 #36 – 44 (even)P. 200 #2 – 22 (even), 21 – 33

(odd), 41, 43, 45

• One way to solve exponential equations is to use the property that if 2 powers with the same base are equal, then their exponents are equal.

• For b > 0 and b≠1if bx = by, then x = y

Solving Exponential Equations

Solve by Equating Exponents

Example 10: Solve 43x = 8x+1

(22)3x = (23)x+1 rewrite with same base

26x = 23x+3

6x = 3x + 3

x = 1Check → 43*1 = 81+1

64 = 64

Your turn!

Solve: 16x = 32x–1

24x = 32x–1

24x = (25)x–1

4x = 5x – 5

x = 5 Be sure to check your answer!!!

When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log

Example 11: Solve 2x = 7

log22x = log27

x = log27

x = ≈ 2.8072log

7log

Example 12: Solve 102x – 3 + 4 = 21

102x – 3 = 17log10102x – 3 = log10172x – 3 = log 172x = 3 + log17x = ½(3 + log17) ≈ 2.115

When you can’t rewrite using the same base, you can solve by taking a log of both sides or use the definition of log

Solve: 5x + 2 + 3 = 25

5x+2 = 22

log55x+2 = log522

x + 2 = log522

x = (log522) – 2

= (log22/log5) – 2

≈ –0.079

Your turn!

Example 13Example 13: Solve ex – 3e-x = 2

More on Solving Exponential Equations

[Answer][Answer] Multiply ex at both sides of the equation:

ex(ex – 3e-x ) = 2ex

e2x – 3 = 2ex

e2x – 2ex – 3 = 0

(ex)2 – 2(ex) – 3 = 0

Denote ex = u, then

(u)2 – 2(u) – 3 = 0

(u)2 – 2(u) – 3 = 0(u – 3)(u + 1) = 0

u = 3, or u = –1

ex = 3, or ex = –1 (discard)

x = ln3

1. To solve use the property for logs with the same base:

• b, x, y+ and b 1• If logb x = logb y, then x = y

2. When you can’t rewrite both sides as logs with the same base exponentiate each side

• b, x+ and b 1• if logb x = y, then x = by • This can get the expression in the log out of the

log simply.

Solving Logarithmic Equations

Example 14: Newton’s Law of Cooling

• The temperature T of a cooling substance at time t (in minutes) is:

• T = (T0 – TR) e-rt + TR

• T0= initial temperature

• TR= room temperature

• r = constant cooling rate of the substance


Example 14: You’re cooking stew. When you take it off the stove the temp. is 212°F. The room temp. is 70°F and the cooling rate of the stew is r = 0.046. How long will it take to cool the stew to a serving temp. of 100°?


T = (T0 – TR) e -rt + TR

T0 = 212, TR = 70, T = 100 r = 0.046So solve:

100 = (212 – 70)e -0.046t + 70

30 = 142e -0.046t (subtract 70)15/71 = e -0.046t (divide by 142)

• How do you get the variable out of the exponent?


ln(15/71) = lne-.046t (take the ln of both sides)ln(15/71) = – 0.046tln(15/71)/(– 0.046) = tt = (ln15 – ln71)/(– 0.046) =t ≈ – 1.556/(– 0.046)t ≈ 33.8 about 34 minutes to cool!

Example 15: Solve log3(5x – 1) = log3(x + 7)

5x – 1 = x + 74x = 8x = 2 and check

log3(52 – 1) = log3(2 + 7)

log39 = log39

• Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions.


Example 16: Solve log5(3x + 1) = –2

3x + 1 = 5-2

3x + 1 = 1/25

x = –8/25 and check

• Because the domain of log functions doesn’t include all reals, you should check for extraneous solutions.


Example 16: Solve log5x + log(x + 1) = 2

log[5x(x + 1)]= 2 (product property)log (5x2 + 5x) = 25x2 + 5x = 100x2 + x – 20 = 0 (subtract 100 and divide by 5)(x + 5)(x – 4) = 0 x = – 5, or x = 4

check and you’ll see x = 4 is the only solution.

More on Solving Logarithmic Equations

log2 [x(x – 7)]= 3

log2 (x2 – 7x) = 3

x2 – 7x = 23

x2 – 7x = 8

x2 – 7x – 8 = 0

(x – 8)(x + 1) = 0

x = 8 x = –1

Your Turn! Solve log2x + log2(x – 7) = 3

Checklog28 + log2(8 – 7) =33 + 0 = 3

log6(x – 2) + log6(x + 3) = 2

log6 [(x – 2)(x + 3)] = 2

log6 (x2 + x – 6) = 2

x2 + x – 6 = 36

x2 + x – 42 = 0

(x – 6)(x + 7)=0

x = 6 x = –7

One More!Solve log6(x – 2) + log6(x + 3) = 2

Checklog64 + log69 =2log636 = 2

Checklog64 + log69 =2log636 = 2

log4 x2 + log4(x – 2)2 = 3 (Raise Power Formula)

log4 [x2(x – 2)2] = 3x2(x – 2)2 = 43 [x(x – 2)]2 = 64 A2B2 = (AB)2

x(x – 2) = ±8 A2 = 64, A = ±8x2 – 2x – 8 = 0 or x2 – 2x + 8 =0x = 4 x = –2 No real solution

Challenge!Example 17: Solve log2x + log4(x2 – 4x + 4) = 3

2log2x + log2(x – 1) = 2

log2x2 + log2(x - 1) = 2

log2 [x2(x – 1)] = 2 x2(x – 1) = 4x3– x2 – 4 = 0(Rational Zero Theorem)(x – 2)(x2 + x + 2) = 0x – 2 = 0 or x2 + x + 2 = 0x = 2 No real solution

Challenge!

Solve log2x + ½ log2(x – 1) = 1

Checklog22 + ½ log21 =12 + 0 = 2

Challenge Simplify (No calculator)

1)

2)

3)

4)

5) Proof

)3(2log32

)5353log(

9106log10)(log 32

3

xdcb dcbaa loglogloglog

2log

1

log

1

52

ππ

Assignment:5.7

P. 201 #24 – 34 (even), 42, 44P. 179 #49 – 52