Math 8

Mathematics homework help

Lesson 4.3

Introduction

Course Objectives

This lesson will address the following course outcomes:

· 24. Translate exponential statements to equivalent logarithmic statements, interpret logarithmic scales, and use logarithms to solve basic exponential equations.

Specific Objectives

Students will understand

· the relationship between a logarithm and an exponent.

Students will be able to

· use logarithms to solve an exponential equation.

In Lesson 4.1 and Lesson 4.2, you created increasing exponential models for compound interest and an exponential decay model for depreciation.  In those types of situations, we might be interested in how long it will take our account to grow to a certain amount, or when the value of a car will drop below a certain level.  To do that, we need to be able to solve exponential equations.

When we solve an equation like 3x=123x=12 , we use division to solve the equation, since division “undoes” multiplication: 3×3=1233×3=123 , so x=4x=4 .  Likewise we use multiplication to “undo” division, addition to “undo” subtraction, and so on.

If we have an exponential equation, it would be helpful to have something to “undo” the exponential so we could solve exponential equations.  The logarithm is that tool.

Common Log

 

Common Logarithm

The common logarithm, written log(x), undoes the exponential 10x.  This means that

log(10x) = x,        and likewise         10log(x) = x.

This also means

the statement 10a = b is equivalent to the statement log(b) = a.

For example, since 102 = 100, then log(100) = 2.

log(x) is read as “log of x”, and means “the logarithm of the value x”.  It is important to note that this is not multiplication – the log doesn’t mean anything by itself, just like √ doesn’t mean anything by itself; it has to be applied to a number.

#1 Points possible: 20. Total attempts: 5

For each of the following, write the number as a power of 10, then use that to evaluate the log.

a) log(1,000) = log(103) =
b) log(100) = log( ) =
c) log(10) = log( ) =
d) log(1) = log( ) =
e) log(0.1) =  log(110)log(110)  = log(10−1)log(10-1) =
f) log(0.01) = log( ) =

It is helpful to note that from the last problem that the number we’re taking the log of has to get 10 times bigger for the log to increase in value by 1.

#2 Points possible: 15. Total attempts: 5

For each of the following, write the number as a power of 10, then use that to evaluate the log.

a) log(100,000) =

b) log(11000)log(11000) =

c) log(0.01) =

Of course, most numbers cannot be written as a nice simple power of 10.  For those numbers, we can evaluate the log using a scientific calculator with a log button. Here a couple videos showing how to use your calculator:

· A graphing calculator (like a TI-84) [+]

· A scientific calculator (like a TI-30x) [+]

#3 Points possible: 5. Total attempts: 5

Evaluate log(150) using a calculator.  Give at least 3 decimal places.

With an equation, just like we can add a number to both sides, multiply both sides by a number, or square both sides, we can also take the logarithm of both sides of the equation and end up with an equivalent equation.  This will allow us to solve some simple equations.

Example:  Solve 10x = 1000:

Taking the log of both sides gives log(10x) = log(1000)

Since the log undoes the exponential, log(10x) = x.  Similarly log(1000) = log(103) = 3.

The equation simplifies then to x = 3.

#4 Points possible: 5. Total attempts: 5

Solve 3(10x)=6,0003(10x)=6,000 .  Give at least 3 decimal places.

This approach allows us to solve exponential equations with powers of 10, but what about problems like 2 = 1.03n from earlier, which have a base of 1.03?  For that, we need the exponent property for logs.

Exponent Property

Properties of Logs: Exponent Property

log(Ar)=rlog(A)log(Ar)=rlog(A)

#5 Points possible: 5. Total attempts: 5

Rewrite each expression using the exponent property of logs.

a. log(1.03x)=log(1.03x)=

b.  log(35x)=log(35x)=

This property will allow us to solve exponential equations. An exponential equation is one where the variable is in an exponent.

Solving exponential equations with logarithms

1. Isolate the exponential.  In other words, get it by itself on one side of the equation.  This usually involves dividing by a number multiplying the exponential.

2. Take the log of both sides of the equation.

3. Use the exponent property of logs to rewrite the exponential with the variable exponent multiplying the logarithm.

4. Divide as needed to solve for the variable.

Example:  Solve 200(1.03x) = 8000:

  200(1.03x) = 8000 Divide both sides by 200
  1.03x = 40 Take the log of both sides
  log(1.03x) = log(40) Use the exponent property of logs on the left
  x log(1.03) = log(40) Divide by log(1.03)
  x=log(40)log(1.03)x=log(40)log(1.03) If appropriate, we can use a calculator to approximate the answer
  x ≈ 124.8  

#6 Points possible: 5. Total attempts: 5

Solve 40(3x)=6,00040(3x)=6,000 .  Give at least 3 decimal places.

#7 Points possible: 5. Total attempts: 5

The value of a car is depreciating following the formula V=18,000(0.82t)V=18,000(0.82t) , where V is the value and t is years since purchase.  Determine when the value of the car will fall to $12,000.  Round your final answer to 2 decimal places (but make sure you keep more during the calculations).

years after purchase

#8 Points possible: 5. Total attempts: 5

The population of a town was 72 thousand in 2010, and has been growing by 8% each year.  When will the population reach 160 thousand if the trend continues?  Give at least 1 decimal place.

years after 2010

Hint: If you have trouble, we’ll provide some hints

HW 4.3

#1 Points possible: 5. Total attempts: 5

Which of the following was one of the main mathematical ideas of the lesson?

· Logarithms are used in carbon dating to determine how old artifacts are

· Logarithms are used to undo exponents

· Logarithmic scales are better than linear scales

· Logarithms can only solve exponential equations where the base of the exponential is 10

#2 Points possible: 5. Total attempts: 5

Evaluate using your calculator, rounding to 3 decimal places:

a. log(71) =

b. log(6814) =

c. log(0.00315) =

d. log(0.72) =

#3 Points possible: 5. Total attempts: 5

Evaluate each of the following, to at least 3 decimal places.

a. log(1,000,000) =

b.  log(11000)log(11000)  =

c. log(70) =

#4 Points possible: 5. Total attempts: 5

Evaluate each of the following, rounding to 3 decimal places if needed:

a. 10001001000100  =

b. log(1000)log(100)log(1000)log(100)  =

Notice the results are different.  This question is to help you notice that log is not something that is multiplying the numbers – it is an operation being applied to the number.  This means you cannot “cancel” the logs.

#5 Points possible: 5. Total attempts: 5

Solve, to 3 decimal places, the equation 2(10x)=4802(10x)=480

x =

#6 Points possible: 5. Total attempts: 5

Solve for k: 200(1.05)k=16,790200(1.05)k=16,790  k=k=      Give an exact answer, or a decimal approximation accurate to at least 3 decimal places.

#7 Points possible: 5. Total attempts: 5

Solve for x. Round your answer to the nearest hundredth. 54x=8954x=89

#8 Points possible: 5. Total attempts: 5

The value of a car is depreciating according to the model V=11000(0.82)tV=11000(0.82)t , where V is the value of the car t years after purchase.

Solve, using logarithms, for when the value of the car will reach $4,000.  Give your answer to the nearest tenth.

years after purchase

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