Lesson 10.1: An Introduction to Logarithms

Learning Goals:

1) What is a logarithm?2) How do we convert between logarithmic and exponential form?

Do Now: Find the inverse of each of the following: switch x∧ y values!

1. f ( x )=3 x+5 2. f ( x )= 3√x+5−3 3. f ( x )=2x

y=3 x+5 y= 3√x+5−3 y=2x

x=3 y+5 x= 3√ y+5−3 x=2y

Solve for y! Solve for y! Solve for y!

x−5=3 y x−3= 3√ y+5

y= x−53

=f−1(x ) ( x−3 )3=( 3√ y+5 )3

( x−3 )3= y+5

y=(x−3)3−5=f −1(x)

What happens when you try to find the inverse of problem 3? Currently we don’t know how to solve for y!

Defining Logarithmic Functions

The function y=logb x is the name we give the inverse of y=bx.

y=logb x is the sameasb y=x

A logarithm gives as its output ( y-value) the exponent we must raise b to in order to produce its input (x-value).

How would you write the inverse to problem 3 using some form of the word logarithm? x=2y so y=log2 x

1

What are logarithms? Logarithms are another way to write an exponential function!

log bN=xmeans bx=N ,where b>0∧b≠1

Parts of a logarithm: base, anti-logarithm, and exponent

log b⏟base⏟

x⏟anti−log

= y⏟exponent

Trick to remember is to “loop”:

log 8 x=13 log x125=−3

2

813 =x x

−32 =125

x=2 (x−32 )

−23=(125)

−23

x= 125

Exponential Form Logarithmic Form x2x=64 log2 64=x 2x=26 x=6

x4=81 log x81=4(x4 )

14=(81)

14 x=±3 so x=+3

52=x log5 x=2 x=25

72=x log7 x=2 x=49

x3=27 log x27=3(x3 )

13=(27)

13 x=3

2

Part 1: Write each exponential equation in logarithmic form.

1. 24=16 2. 813 =2

log2 16=4 log 82=13

Part 2: Write each logarithmic equation in exponential form.

3. log 432=52 4. log a

1a=−1

452=32 a−1=1

a

Part 3: Solve each equation for x:

5. log2 x=4 6. log 81 x=12

24=x 8112=x

x=16 x=9

7. log6 x=−2 8. log64 x=23

6−2=x 6423=x

x= 136 x=16

Part 4: Evaluate each expression:

9. log25 125=x set it ¿ x 10. log51

25=x set it ¿ x

25x=125 5x= 125

(52 )x=53 5x=25−1

2 x=3 5x=(52 )−1

x=32 x=−2

3

11. log 4917=x set it ¿ x 12. log100 0.1=x set it ¿ x

49x=17 100x=0.1

49x=7−1 100x= 110

(72)x=7−1 100x=10−1

2 x=−1 (102 )x=10−1

x=−12 2 x=−1

x=−12

Part 5: Solve each equation for b.

13. log b27=3 14. log b12=−1

b3=27 b−1=12

3√b3=3√27 (b−1 )−1=( 1

2 )−1

b=3 b=2

15. log b√5= 14 16. log b 4=1

2

b14=√5 b

12 =4

(b14 )4

=(512 )4

(b12)2

=( 4 )2

b=52=25 b=16

17. If f ( x )=log3 x, find f (1)

f (1 )means x=1!

4

f (1 )=log3 1← logbase in calc!

y=log31

3y=1

y=0

18. Verify the following by evaluating the logarithms:

log2 (8 )⏟+ log2 ( 4 )⏟=log2(32)⏟ log2 (8 )⏟+ log2 ( 4 )⏟=log2(32)⏟

2x=8 2x=4 2x=32 3+2=5

2x=23 2x=22 2x=25 5=5√

x=3 x=2 x=5

5

Homework 10.1: An Introduction to Logarithms

Write each of the following in logarithmic form.

1. 216=63 2. 4−2=0.0625 3. 62534 =125

Write each of the following in exponential form.

4. 5=log3 243 5. −2= log5 0.04 6. log 49343=32

Evaluate each of the following:

7. log6 216 8. log2 32 9. 5 ∙ log8 8

Solve each equation for the variable.

10. a=log 416 11. log 8 x=12 12. log b64=6

6

13. If f ( x )= log3 x, find f (81).

14. When $1 is invested at 6 % interest, its value, A, after tyears is A=1.06t. Express t in terms of A.

15. Can the value of log2(−4) be found? What about the value of log2 0? Why or why not? What does this tell you about the domain of log b x?

16. Which of the following is equivalent to y=log7 x?

(1) y=x7 (2) x= y7 (3) x=7 y (4) y=x17

17. Verify the following by evaluating the logarithms:

log 4 (4 )+ log4 (16 )=log4 (64)

7

Lesson 10.2: Natural Logarithms

Learning Goals:

1) What is a logarithm?2) What is the number e?3) What is a natural logarithm?4) What is a common logarithm?5) How do we convert between logarithmic and exponential form?

Do Now: Which of the following are valid examples of logarithms?

log2 4 log1 4log 0 4 log3 27log5(−25) log7 0Valid Examples of Logarithms Invalid Examples of Logarithms

log3 27

log2 4

log 0 4

log5(−25)

log1 4

log7 0

Definition of the Logarithm base b:

If three numbers, L ,b ,∧x with 0<b<1∨b>1 are related by bL=x, then L is the logarithm base b of x, and we write log b ( x )=L.

8

Compound Interest: You deposit $3500 in an account that pays 4 % annual interest compounded continuously. What is the balance after 1 year?

Solution: P=3500 r=.04 t=1

P=amount you invest ( principal ) r=interest rate (%100 )t=time

Use the formula for continuously compounded interest.

A=Per t Write the formula

¿3500 e( .04 ∙1 ) Substitute for P , r ,∧t

A=3642.84 The balance at the end of 1 year is$3642.84

Complete the following example:

You deposit $4800 in an account that pays 6.5 % annual interest compounded continuously. What is the balance after 3 years?

When it says compounded continuously use base of e!

Solution: P=4800 r=.065 t=3

P=amount you invest ( principal ) r=interest rate (%100 )t=time

Use the formula for continuously compounded interest.

9

The Natural Base e

The natural base e is irrational. It is defined as follows:

As n approaches +∞,(1+ 1n )

n

approaches

e ≈2.718. 2nd→÷→e

Natural Base Functions

A function of the form y=aer x is called a natural base exponential function.

If a>0 and r>0, the function is an exponential GROWTH function. If a>0 and r<0, the function is an exponential DECAY function.

A=Per t Write the formula

¿4800 e(.065 ∙3 ) Substitute for P , r ,∧t

A=5833.49 The balance at the end of 3 years is$5833.49

The natural logarithms, gives an exponent as its output. In fact, it gives the power that we must raise e to in order to get the input.

y=ln x ln x=loge x ln x ≠ lne x

Without using your calculator, determine the values of each of the following.

1. ln (e )=x 2. ln 1=x

log ee=x log e1=x

ex=e1 ex=1

x=1 x=0

3. ln (e5 )=x 4. ln √e=x

log ee5=x log e√e=x

ex=e5 ex=√e

x=5 x=12

Rewrite each of the following in logarithmic form.

5.. e5=x 6. ex=15

log e x=5 log e15=x

ln x=5 ln 15=x

10

The Natural Logarithm

The inverse of y=ex : y=ln x y=loge x

Common and Natural Logarithms {Calculator Information}

Common Logarithm: A logarithm with a base of 10.

On the graphing calculator, the Log key is used to display the common logarithm of a number.

Examples: Find the common log of each number to four decimal places.

1. 279=log (279 )=2.4456 2. 0.0005=log ( .0005 )=−3.3010

When I find the common log of a number, it means that

102.4456≈279∧10−3.3010≈0.0005

How can we find the number if we know the common log of the number?

log ( x )= log10(x )

Examples: Find to the nearest thousandth each number N whose common logarithm is given.

3. log N=3.9294 4. log N=−1.7799

log10 N=3.9294 log10 N=−1.7799

103.9294=N 10−1.779=N11

On the graphing calculator, the ln key is used to display the natural logarithm of a number.

Find the natural log of each number to four decimal places.

5. ln 52=3.9512 6. 2 ln 16=5.5452

When I find the natural log of a number, it means that e3.9512≈52

How can we find the number if we know the natural log of the number?

ln ( x )=loge(x )

Examples: Find to four decimal places the antilogarithm of the given logarithm.

7. ln x=−.5373 8. ln x=.8297

e−.5373=x e .8297=x

x=.5843 x=2.2926

Compute the value of each logarithm. Verify your answers using an exponential statement.

1. log13 (13 )=1 2. log √7( 149 )=−4

13x=13 √7x= 149

13x=131 (7 12)x=49−1

x=1 (7 12)x=(72 )−1

12x=−2

x=−4

12

What is wrong with the following questions?

3. log2(0) 4. log3(−13 )

They can’t be evaluated!

Find the value of each of the following.

5. If x=log2(8) and y=2x, 6. If r=26 and s=log2(r ),

find the value of y. find the value of s.

2x=8∧ y=2x , then y=8 2s=r sub in the value of r!

2s=26

s=6

Rewrite each of the following in exponential form.

7. ln 25=x 8. ln x=52 9. log 45=x 10. log x=.32

ex=25 e52=x 10x=45 10x=.32

Find x using your calculator. Round all answers to the nearest hundredth.

11. ln 45=x 12. log 75=x 13. ln x=3.2958 14. log x=−.4881

x=3.81 x=1.88 x=27.00 x=0.33

15. Solve for x :3 log ( x+4 )=6 16. Solve for x to the nearest

3 log (x+4 )3

=63 thousandth:

log ( x+4 )=2 4+ ln ( x−1 )=4.8200

102=x+4 ln ( x−1 )=.82

100= x+4 e .82=x−1

x=96 x=e .82+1

You can always check your answers! x=3.270

13

Homework 10.2: Natural Logarithms

1. When Kyle was born, his grandparents invested $5000 in a college fund that paid 4 % compounded continuously. What is the value of the account after 18 years? {A=Per t }

2. A hot liquid is cooling in a room whose temperature is constant. Its temperature can be modeled using the exponential function shown below. The temperature, T , is in degrees Fahrenheit and it is a function of the number of minutes, m, it has been cooling.

T (m )=101e−0.03m+67

a) What was the initial temperature of the water at m=0?b) How do you interpret the statement that T (60 )=83.7?

14

Determine the value for each of the following logarithms.

3. log2( 164 ) 4. log 4 8 5. log2

5√4

6. ln (e4 ) 7. ln 3√e 8. ln e

Rewrite in logarithmic form.

9. e7=x 10. ex=7 11. 2x=5

12. log 927 13. log x (x3 )

Find x in each of the following. Round answers to the nearest hundredth.

14. ln x=3.3534 15. log x=1.7218

16. log 0.528=x 17. ln 51.3=x

15

Lesson 10.3: Properties of logarithms

Learning Goal: How do we use the properties to expand and simplify log expressions?

Do Now: Answer the following questions in order to prepare for today’s lesson.

1. Simplify: (3 x5 y4 )2 ∙ (2 s4 y )6 x5 y6 =

(32∙ x10 ∙ y8 ) (2 x4 y )6x5 y6 =18x14 y9

6 x5 y6 =3 x9 y3

(1) Power Law (2) Product Law (3) Quotient Law

2. Write each expression in radical form.

a. 712 index ¿2 b. 2

53 index ¿3 c. (3m )

74 index ¿4

√7 3√25 4√ (3m )7

3. Write each expression in exponential form.

a. 3√2=213 b. (√10 )3=10

32 c. 4√ x5 y=x

54 y

14

4. What are logarithms? Logs are exponents!

5. Label the base, exponent, and anti-logarithm: log bN=x

16

Using the properties of exponents, we can arrive at the properties of logarithms.

Properties of Exponents Properties of LogarithmsMultiplication Law

bn∙ bm=bn+m log b (m ∙n )=logbm+logbn

Division Law bn

bm=bn−m log b(mn )=logbm−logbn

Power Law (bn )m=bn ∙m log bmr=r logbm

When separating logs, the power law is done last!

Directions: Rewrite each expression as sums and differences in terms of log ( x ) , log ( y ) ,∧log (z )

1. log x3

y=3 log x−log y

2. log bx √ y=logb x+ logb√ y=logb x+ logb y12=logb x+

12

logb y

3. log xyz

=log x−( log y+log z )=log x−log y−log z

4. log (√x7 y3 )=log (√7 ∙√ y3 )=log√7+ log√ y3=log712+ log y

32

¿ 12

log 7+ 32

log y∨12(7 log x+3 log y)

5. log 3√ xy6 = log x

13− log y2=1

3log x−2 log y

6. log(3√x y7

x2 z )=log (x13)+log y

73−log x2−log z=¿

13

log x+73

log y−2 log x−log z

17

Directions: Use the properties of logarithms to rewrite each expression in an equivalent form containing a single logarithm.

1. 2 log5m+ log5n= log5m2+log5n=log5(m

2n)

2. 12

logm−3 log n=log √m−log n3=log(√mn3 )3. log 2+ log I−logT=log(2 I

T )4. 2 log x−(3 log y+log z )=log x2−( log y3+log z )=log( x2

y3 z )5. 2 log x+3 log y−1

2log z=log x2+ log y3−log √z=log( x2 y3

√ z )6. 1

3 ( log ( x )−3 log ( y )+ log (z))=3√ log( xy3 z )

When proving equations, you must get the left side ¿ to the right side!

Directions: Use properties of logarithms to show that:

1. log (26 )−log (13 )=log (2) 2. log (3 )+ log ( 4 )+ log (5 )−log (6 )=1

log( 2613 )=log2 log( 3 ( 4 ) (5 )

6 )=1

log 2= log 2√ log 10=1

1=1√

3. 12

log (25 )+ log ( 4 )=log (20) 4. log( 12−1

3 )+ log (2 )=log ( 13 )

log √25+log 4=log 20 log( 16 )+log 2=log( 1

3 ) log 5+ log 4=log 20 log( 1

6∙2)=log( 1

3 ) log (5 ∙4 )=log 20 log( 1

3 )=log(13 )√

log 20=log 20√

18

5. log ( 13− 1

4 )+(log ( 13 )−log( 1

4 ))= log( 19 )

log( 112 )+ log(

1314

)=log( 19 )

log( 112 )+ log( 4

3 )=log( 19 )

log( 112

∙ 43 )=log( 1

9 ) log( 1

9 )= log( 19 )√

19

Homework 10.3: Properties of logarithms

1. The expression log 4 x is equivalent to:

(1) log x4 (2) 4 log x (3) log 4+ log x (4) ( log 4)(log x)

2. log √xyz is equal to:

(1) 12

log x+ 12

log y−log z (2) 12

log x+ log y−log z

(3) 12( log x+ log y−log z) (4)

12

log xy

log z

3. The expression 3 log x−12

log y is equal to

(1) log x3

y2 (2) log x3

√ y (3) log √ 3 x

y (4)

log 3 x12

log y

4. Apply properties of lograrithms to rewrite the following expressions as a single logarithm or number.

a. 12

ln (36 )+ ln (2)

b. ln ( 4 )−3 ln( 13 )+ ln (2)

c. ln (5 )+ 35

ln (32 )−ln (4)

5. Apply properties of logarithms to rewrite each expression as a sum of terms involving numbers, log ( x ) , log ( y ) ,∧log (z ), where x , y ,∧z are positive real numbers.

a. log (xy z3 ) b. log( 3√ x3

yz ) c. log (3 x4 √ y5 )

6. Use properties of logarithms to show that:20

a. log (2 )−log( 113 )=log (26) b. 1

2log (16 )+ log (3 )+ log( 1

4 )=log (3)

Lesson 10.4: More Properties of Logarithms and Substitution

Learning Goals:

1) How do we use the properties to expand and simplify log expressions?2) How do we simplify a logarithmic expression using powers of 10 rules?

Do Now: Answer the following questions in order to prepare for today’s lesson.

1. What is a common logarithm? How do you know when you have a common logarithm? A common log has no base visible so we know base ¿10 for example log x=log10 x

2. Complete the following table of logarithms by using a calculator; then, answer the questions that follow.

x log (x)1,000,000 106=6100,000 105=510,000 104=41,000 103=3100 102=210 101=1

a) What is log (1)? How does that follow from the definition of a base-10 logarithm?

log (1 )=0100=1count ¿of zeros sincethe base=10

b) What is log (10k) for an integer k? How does that follow from the definition of a base- 10 logarithm?

Write a conjecture using logarithmic notation.

log (10k)=k k is the¿ of zeros ! 21

Logarithm Rules from Previous Lesson

Multiplication Lawlog b (mn )=logbm+ logbn

Division Lawlog b(mn )=logbm−logbn

Power Lawlog bm

r=r logbm

New Logarithm Rules

When expanding logarithms, you must be aware of special logs that can be reduced!

Below are two special cases:

log (10k)=k ln (e )=1

k is the # of zeros or exponent of 10. ln (e )= logee=1 so e1=e

Directions: Apply properties of logarithms to rewrite each expression as a sum of terms involving numbers, log ( x ) , log ( y ) ,∧log (z ), where x , y ,∧z are positive real numbers.

a. log (1000√ x )=log 1000+log √ x=3+ 12

log x

b. log( 100 x2

y3 )= log100+2 log x−3 log y=2+2 log x−3 log y

c. log( 110 x2 z )=log 1−(log 10+2 log x+ log z )=0− (1+2 log x+log z )

¿−1−2 log x−log z

Directions: Write each expression as a sum or difference of constants and logarithms of simpler terms. Remember: ln e=1

a. ln (3e )= ln3+ ln e=ln 3+1

b. ln ( e4

xy )=ln e4−( ln x+ ln y )=4 ln e−ln x−ln y=4−ln x−ln y

c. ln (√5 x3

e2 )= ln (√5 x3 )−ln e2=ln (512 ∙ x

32)−ln e2=¿

22

12

ln 5+ 32

ln x−2 ln e=12

ln5+ 32

ln x−2∨12

( ln 5+3 ln x )−2

Substitution with Logarithms

When asked to substitute using logarithms, you must EXPAND the logarithms first by using the properties of logarithms.

Example 1: If log 2=x∧log 3= y, express log 6 in terms of x∧ y.

log 6=log (2∙3 )=log 2+log 3=x+ y

Example 2: If ln 2=x∧ln 3= y, express ln 49 in terms of x∧ y.

ln 4−ln 9=ln 22−ln 32=2 ln 2−2 ln3=2x−2 y

Example 3: If log 5=a, then log 250 can be expressed as:

(1) 50a (2) 2a+1 (3) 10+2a (4) 25a

log 250=log (10∙25 )=log 10+log 25=1+ log52=1+2 log5=1+2a

Example 4: If log x=a , log y=b ,∧log z=c, rewrite log x2 y√z in terms of a ,b ,∧c .

log x2+ log y− log√ z=2 log x+ log y−12

log z=2a+b−12c

Practice

1. Use the approximate logarithm values below to estimate the value of each of the following logarithms.

log (2 )=0.3010

log (3 )=0.4771

log (7 )=0.8451

a. log 12=log (4 ∙3 )=log (22∙3 )=2 log 2+log 3=2 ( .3010 )+.4771=1.0791

b. log( 19 )=log1−log 9=0−log32=−2 log 3=−2 ( .4771 )=−.9542

c. log (70 )=log (7 ∙10 )=log 7+ log10=.8451+1=1.8451

23

2. Given: log b3=p∧logb5=q

Express each of the following in terms of p∧q .

a. log b 45=logb ( 9∙5 )=logb (32 )+log b5=2 logb3+ logb 5=2 p+q

b. log b√ 35=logb

√3√5

= logb√3−logb√5=12

logb3−12

logb 5=¿

12p−1

2q∨1

2( p−q)

c. log b273√5

=logb 27−logb3√5=log b (33 )−log b

(513 )=¿

3 logb 3−13

logb 5=3 p−13q

d. log b9

3√5=logb 9−logb

3√5=log b (32)−logb(5

13 )=¿

2 logb 3−13

logb5=2 p−13q

3. If ln a=2∧ln b=3, what is the numerical value of ln ( e√ab3 )?ln e+ 1

2ln a−3 ln b=1+ 1

2(2 )−3 (3 )=1+1−9=−7

24

Homework 10.4: More Properties of Logarithms and Substitution

1. If log a=x∧logb= y, then log √ab is equivalent to:

(1) 12x+ y (2) 1

2(x+ y ) (3) 1

2xy (4) 1

4xy

2. Given: ln 2=x∧ln 3= y

Express each of the following in terms of x∧ y : ln √29

3. Express each of the following as a single logarithm.

(a) log (8 )+ log( 14 )

(b) 13

log (8 )+2 log (3 )−log (6)

(c) 3 log ( x )−(2 log ( y )+ 12

log (z))

4. In the following expression, x , y ,∧z represent positive real numbers. Use properties of logarithms to rewrite the expression in an equivalent form containing only log ( x ) , log ( y ) , log ( z ), and numbers.

log √( x3 y2

10 z )

5. Rewrite the expression as sums and differences in terms of ln ( x ) , ln ( y ) ,∧ln (z ).

25

ln ( ey z3 )

Lesson 10.5: Solving Exponential Equations Using Logarithms

Learning Goal: How do we use logarithms to solve an exponential equation?

a) Solve the following equation for x : 43 x−1=32

Get like bases! (22 )3 x−1=25

2 (3x−1 )=5

6 x−2=5

6 x=7

You can always check your answer! x=76

b) What happens when you try to solve 52 x=20 for x?

You cannot get like bases, so we will need to use logs!

How can we solve exponential equations when you cannot get like bases?

1) Look to isolate the base with the variable exponent.2) Take the common log of both sides3) Use the power law to move the exponent4) Solve for the variable

1. 2 ∙52 x=20

2∙52x

2=20

2

52 x=10

2 x log 5=log 10

2x log 5log 5

= log10log5

2 x=1.430676558

26

x=.715338279

Solve each equation for the given variable and express your answer to the nearest hundredth.

1. 72w+8=34 2. e2 x=5

72w=26 2 x ln e=ln5∨2 x log e=log5

2w log 7=log 26 2 x=ln 5∨2 x= log 5log e

2w log 7log7

= log 26log 7 x=.80

2w=1.67433041

w=0.84

3. 52+ x−5x=10 4. 32 x−3=2x+4

GCF: 5x (2 x−3) log3=(x+4) log2

5x (52−1 )=10 2 x−3=(x+4) log 2log 3

5x (25−1 )=10 2 x−3=(x+4)(.6309297536)

5x (24 )=10 2 x−3=.63 x+2.52

5x= 512 1.37 x=5.52

x log5=log 512 x=4.03

x=−0.54

Solve each equation for the given variable and express your answer to the nearest hundredth.

5. 5e3x+ 4=15 6. 5x+3=42 x−1

e3x+4=3 ( x+3 ) log 5=(2 x−1 ) log 4

(3 x+4 ) ln e=ln 3 x+3=(2 x−1) log 4log 5

3 x+4=ln 3 ( x+3 )= (2 x−1 )(.8613531161)

27

x= ln (3)−43

x+3=1.72 x−.86

x=−.9671292371 −.72 x=−3.86

x=−0.97 x=5.36

7. 2 (4x )+4x+1=342 8. √3 ∙33 x=9

GCF: 4 x3

12 ∙33 x=32

4 x (2+41 )=342 312+3x

=32

4 x (6 )=342 12+3 x=2

4 x=57 3 x=32

x log 4=log 57 x=12

x=2.92

Applications of Logarithms

9. You are investing $600 at an interest rate of 7.5% compounded continuously. Using the formula, A=Per t, where A is your final amount, P, is your starting amount, r is the interest rate, and t is the time in years, determine how long it will take you to reach $1000, to the nearest tenth of a year?

1000=600e.075 ∙t 50=80(0.98)t,

53=e .075∙ t 5

8=(0.98)t

ln (53 )=.075 t (ln e) log( 5

8 )=t log (0.98)

t=6.8 t=23.26438837

1998+23=2021

10. A population of wolves in a county is represented by the equation

P ( t )=80(0.98)t, where t is the number of years since 1998. In what year will 50 wolves be reached? See above to right for answer

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Homework 10.5: Solving Exponential Equations Using Logarithms

1. Solve the following exponential equation and express your answer to the

nearest hundredth: ( 12 )

x3 +1

=16 .

2. Solve the following equation for x and express your answer to three decimal places:

2x+1=31− x

3. Find the solution to the exponential equation below to the nearest hundredth:

4 (2)x−3=17

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4. The logarithmic expression ln (√ey3 ) can be rewritten:

(1) 3 ln y−2 (2) 1−6 ln y2 (3) ln y−6

2 (4) √ ln y−3

5. The savings bank account can be modeled using S (t )=1250 e0.045 t, where t is the number of years the money has been in the account. Determine, to the nearest tenth of a year, how long it will take for the amount of savings to double from the initial amount deposited of $1250.

6. Matt places $1200 in an investment account earning an annual rate of 6.5 %, compounded continuously. Using the formula A=Per t, how long will it take for his initial investment to quadruple in value? Round to the nearest year.

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Lesson 10.6: Solving Logarithmic Equations

Learning Goals:

1) How do we solve logarithmic equations?2) How do we check for extraneous roots?

Warm Up:

Using what we have learned about solving for an exponent, answer each of the following questions:

a) Solve the following equation for t : a∙bct=d

a ∙bc t

a=d

a

bct=da

ct (log b )=log( da )

c t (log b )c (log b)

=log( da )c (log b)

t=log( da )c (log b)

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RECALL

bL=x can be written in the form log b ( x )=L where

0<b<1∨b>1∧x>0

*This means the base cannot be 0, 1, or a negative number and the

anti-logarithm cannot be 0 or a negative number.

b) Solve the following equation for a :3+xa+b=c

xa+ b=c−3

(a+b ) log x=log (c−3)

a+b= log (c−3)log x

a= log (c−3)log x

−b

There are 3 types of log equations:

Type Procedure Example1 log “loop”

Cannot loop until you isolate!log ((2x+5)2 )=4

104=(2x+5)2

3−log2 (x−1 )=0−log2 ( x−1 )=−3

log2 ( x−1 )=323=x−1

12

log ( x+2 )=2

log ( x+2 )12=2

log ( x+2 )12 ∙ 2

=22

log ( x+2 )=4104=x+2

1 log on each side

“cancel”Logs must have like bases!

2 ln ( x+2 )=ln (−x)ln (x+2)2=ln (−x )

(x+2)2=−x2 or more logs on one side

“combine” logm+log n=log (m∙n)

logm−log n=log (mn ) log2 (3 x )+ log2 ( 4 )=4

log2 (12x )=424=12x

2 log5 x−log5 5=log5125log5 x

2−log5=log5 125x2

5=125

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Practice! Determine which of the three methods would be most appropriate before solving each problem. Be sure to check for extraneous solutions.

1. log2 (9 x2+30 x+25 )=8 2. ln ( x+2 )= ln (12 )−ln ( x+3 )

28=9 x2+30 x+25 x+2= 12x+3 Cross Multiply!

256=9 x2+3 0 x+25 x (x+2 ) ( x+3 )=12

0=9 x2+30 x−231 x2+5x+6=12

0=3(3 x2+10x−77) x2+5x−6=0

Quadratic Formula or AC Method ( x−1 ) ( x+6 )=0

x=113

∧−7√ x=1∧−6

Always check! Omit −6 so the only answer is x=1

x+2>0 x+3>0

−6+2>0 −6+3>0

−4>0 no −3>0 no

3. log (x2+7 x+12 )−log ( x+4 )=0 4. log7 ( x−2 )+ log7 ( x+3 )=log7 14

x2+7 x+12x+4

=100 ( x−2 ) ( x+3 )=14

(x+4 )(x+3)(x+4)

=1 x2+ x−6=14

x+3=1 x2+ x−20=0

x=−2√ ( x−4 ) ( x+5 )=0

Always check! x=4∧x=−5(omit after checking)

5. ln ( (4 x )5 )=15

5 ln (4 x)=15

ln 4 x=3

e3=4 x x=5.021384231√ Always check!

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6. log2 ((5 x+7 )19 )=57

19 log2 (5x+7 )=57

log2 (5 x+7 )=3

5 x+7=23

5 x+7=8

5 x=1

x=15√ Always check!

7. log(x+ 2)(x2+5 x+18 )− log ( x+2) (2 )=2

x2+5 x+182

=(x+2)2

x2+5 x+182

=(x+2)(x+2)

x2+5 x+182

=x2+4 x+4

2 x2+8 x+8=x2+5 x+18

x2+3x−10=0

( x−2 ) ( x+5 )=0

x=2∧−5 Always Check!

Omit −5 so only x=2 will work!

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Homework 10.6: Solving Logarithmic Equations

1. Solve for x : ln (x+2 )+ ln ( x−2 )=ln (9 x−24 )

2. Solve for x : log (10 x+5 )−3= log (x−5)

3. Solve for x : ln ( 32x2 )−3 ln (2 )=3

4. Solve for x : log2 ( x )+log2 (2x )+ log2 (3 x )+ log2 (36 )=6

5. Jenn claims that because log (1 )+log (2 )+log (3 )=log (6), then

log (2 )+ log (3 )+ log ( 4 )=log (9). Is she correct? Explain how you know.

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Lesson 10.7: Change of Base Rule

Learning Goals:

1) What is the change of base rule for logarithms?2) How do we solve logarithmic equations using change of base rule?

Warm Up:

Evaluate the following:

a) log2 8 use “logbase” in calc b) log 418

2x=8 4 x=18

x log2=log 8 x log 4=log 18

x= log 8log 2

=3 x= log 18log 4

=2.1

1.

Let’s look at the two examples from the warm-up. Use the change of base rule to rewrite the logarithms with base10 and then use the LOG key in your calculator.

a) log2 8= log8log2

=3 b) log 418= log 18log 4

=2.1

2. Use the change of base property to rewrite each logarithmic function in terms of the common logarithm function.

a) g1 ( x)=log 14

( x) b) g2 ( x)=log 12

(x) c) g3 ( x )=log2(x )

log x

log 14

log x

log 12

log xlog 2

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Change of Base Formula for Logarithms

If x ,a ,∧b are all positive real numbers with a≠1∧b≠1, then

log b ( x )=loga( x)log a(b)

3. Using the change of base rule, rewrite the given logarithmic expressions in base 7.

a) log2 11 b. log5 x2 c. log b(x+3)

log7 11log72

log7 x2

log75log7(x+3)

log7b

Push yourself!

Show that for any positive numbers a∧bwith a≠1∧b≠1,

log a (b ) ∙ logb (a )=1.

logbloga

∙ log alog b

=1

1=1√

How can we use the change of base formula to solve a logarithmic equation?

Rewrite the logarithmic expresssion with the larger base in terms of the smaller base.

Cross multiply and drop the logs on both sides.

1. Solve for x : log (x )=log100 (x2−2 x+6 )

log x1

=log (x2−2x+6)

log100

log x1

= log (x2−2x+6)2

2 log x=log (x2−2x+6)

log x2=log (x2−2x+6) Cancel like bases!

x2=x2−2 x+6

2 x=6

x=3√ Always check! log ( x )=log100 (x2−2x+6 )

x>0 (x2−2 x+6 )>0

3>0 9>0

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2. Solve for x : log2 ( x+1 )=log4⏟( x2+3 x+4)

Change to base 2!

log2 ( x+1 )=log2(x

2+3x+4)log2 4

log2 ( x+1 )=log2(x

2+3x+4)2

2 log2 (x+1 )=log2(x2+3x+4)

log2 ( x+1 )2=log2(x2+3 x+4) Cancel like bases!

( x+1 )2=(x2+3 x+4)

( x+1 ) ( x+1 )=x2+3 x+4

x2+2x+1=x2+3 x+4

−x=3

x=−3 Reject so “No Solution”!

3. Solve for x : log9⏟ (x2+2 x+6 )=log3(x+2)

Change to base 3!

log3(x2+2x+6)

log39=log3(x+2)

log3(x2+2x+6)2

=log3(x+2)

2 log3 ( x+2 )=log3(x2+2x+6)

log3 ( x+2 )2=log3(x2+2 x+6) Cancel like bases!

(x+2)2=x2+2 x+6

( x+2 ) (x+2 )=x2+2 x+6

x2+4 x+4=x2+2 x+6

2 x=2 Always check! log 9 (x2+2 x+6 )= log3 ( x+2 )

x=1√ x2+2x+6>0 x+2>038

9>0 3>0

4. Solve for x : log (x−2 )=log100⏟(14−x )

Change to base 10!

log ( x−2 )= log (14−x)log 100

log ( x−2 )= log (14−x)2

2 log ( x−2 )= log (14−x)

log( x−2)2=log (14−x ) Cancel like bases!

(x−2)2=14−x

( x−2 ) ( x−2 )=14−x

x2−4 x+4=14−x

x2−3 x−10=0

( x−5 ) ( x+2 )=0

x=5∧−2 Always check! log ( x−2 )=log100 (14−x)

x−2>0 14−x>0

Must omit the −2 −4>0 16>0

So only x=5√ x−2>0 14−x>0

3>0 9>0

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Homework 10.7: Change of Base Rule

1. Use the change of base rule to solve for x in the given logarithmic equation:

log ( x+1 )=log100 (x2+4 x )

2. Solve the given logarithmic equation: log 4 (x )=log16(3 x2−3 x+1)

3. Which expression could be used to determine the value of y in the equation

log6 x= y? (1) log x6 (2) log 6

log x (3) 6log x (4) log x

log 6

4. Solve for x : ln (x+2 )+ ln ( x−2 )=ln (−2x−1)

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5. Solve for x : log (√ ( x+3 )3)=32

6. Drew said that the equation log2 [ ( x+1 )4 ]=8 cannot be solved because he expanded (x+1)4=x4+4 x3+6 x2+4 x+1 and realized that he cannot solve the equation x4+4 x3+6x2+4 x+1=28. Is he correct? Explain how you know.

7. Write ln (√5 x3

e2 ) as a sum or difference of constants and logarithms of simpler

terms.

41