precalculus chapter 3 practice test name:
TRANSCRIPT
PreCalculusChapter3PracticeTest Name:____________________________
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A note on terminology: Zeros and roots are the same thing. If they are real, as opposed to complex, they are also
š„āintercepts of your graph.
In the log expression, log 64 2, first is ābā, last is ā2ā and middle is ā64.ā We put these in an
exponential expression, from left to right, to get: šš šš.
PreCalculusChapter3PracticeTest Name:____________________________
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In the log expression, log 216 š„, first is ā6ā, last is āš„ā and middle is ā216.ā We put these in
an exponential expression, from left to right, to get: šš ššš.
Firstālastāmiddle works this way too.
In the exponential expression, 2 š„, first is ā2ā, last is āš„ā and middle is ā3.ā We put these in a
logarithmic expression, from left to right, to get: š„šØš š š š.
In the exponential expression, 2 , first is ā2ā, last is ā ā and middle is ā 2.ā We put these
in a logarithmic expression, from left to right, to get: š„šØš šš
šš.
In the log expression, log 10 š„, first is ā10ā, last is āš„ā and middle is ā10.ā We put these in
an exponential expression, from left to right, to get: 10 10, then solve:
log 10 š„ converts to: 10 10 š š
In the log expression, log ā3 š„, first is ā3ā, last is āš„ā and middle is āā3.ā We put these in an
exponential expression, from left to right, to get: 3 ā3, then solve:
log ā3 š„ converts to: 3 ā3 š š
š
Note: ā3 3
PreCalculusChapter3PracticeTest Name:____________________________
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In the log expression, log 1 š„, first is ā6ā, last is āš„ā and middle is ā1.ā We put these in an
exponential expression, from left to right, to get: 6 1, then solve:
log 1 š„ converts to: 6 1 š š
Problems 8 to 9: Exponentiation and taking logarithms are inverse operations, so when they both exist,
with the same base, they cancel each other out.
š š„šØš š šš šš
š„šØš š ššš šš
Back to āfirstālastāmiddle.ā
In the log expression, log 1000 š„, first is ā10ā, last is āš„ā and middle is ā1000.ā We
put these in an exponential expression, from left to right, to get: 10 1000, then solve:
log 1000 š„ converts to: 10 1000 š š
In the log expression, log 10 š„, first is ā10ā, last is āš„ā and middle is ā10 .ā We put these
in an exponential expression, from left to right, to get: 10 10 , then solve:
log 10 š„ converts to: 10 10 š š
Note that ln š„ is equivalent to log š„. Then,
In the log expression, log š š„, first is āeā, last is āš„ā and middle is āš.ā We put these in
an exponential expression, from left to right, to get: š š, then solve:
log š š„ converts to: š š š š
Note: I like to use parentheses to make it
easier to read a problem. For example,
Problems 8 and 9 have values at multiple
levels. By using parentheses, I can make it
easier to see what is going on in the problem.
PreCalculusChapter3PracticeTest Name:____________________________
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log 8š„ log 8 log š„ š š„šØš š š
log125
š„log 125 log š„ š š„šØš š š
log š¦š§ log š¦ log š§ š„šØš š š š š„šØš š š
logš„ 6
š„log š„ 6 log š„ š„šØš š š š š š„šØš š š
log š„ 8 log š„ 4 š„šØš šš šš š
3 log š„ 5 log š„ 6 log š„ log š„ 6 š„šØš š šš š š š
4 log 2 log 8 log 2 log 2 log22
š„šØš š š
Alternatively,
4 log 2 log 8 4 log 2 log 2 4 log 2 3 log 2 š„šØš š š
PreCalculusChapter3PracticeTest Name:____________________________
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Starting equation: 4 64
Take the "log " of both sides: 1 2x 3
Subtract 1: 2š„ 2
Divide by 2: š š
Starting equation: š
Convert exponent on the right: š š
Take the "ln" of both sides: x 8 4
Subtract 8: š šš
Starting equation: 5 3
Take the "ln" of both sides: ln 5 ln 3
Simplify: x 7 ln 5 ln 3
Divide by ln 5: x 7
Subtract 7: š š„š§ šš„š§ š š
Note: I tend to use ln in
solving problems like this.
However, you can use logs
with any base. Other useful
bases for this problem might
be log, log3 or log5.
PreCalculusChapter3PracticeTest Name:____________________________
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Starting equation: š 2
Take the ln of both sides: ln š ln 2
Simplify: x 4 ln 2
Subtract 4: š š„š§ š š
Starting equation: log š„ 1 1
Take 3 to the power of both sides: 3 3
Simplify: x 1
Add 1: š š
š
Starting equation: 4 8 ln š„ 8
Subtract 4: 8 ln š„ 4
Divide by 8: ln š„
Take š to the power of both sides: š š
Simplify: š± šš
š āš
PreCalculusChapter3PracticeTest Name:____________________________
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Starting equation: log š„ log š„ 35 2
Combine log terms: log š„ ā š„ 35 2
Take 6 to the power of both sides: 6 ā 6
Simplify: š„ ā š„ 35 36
Distribute š„: š„ 35š„ 36
Subtract 36: š„ 35š„ 36 0
Factor: š„ 36 š„ 1 0
Determine solutions for š„: š„ 36, 1
Test the solutions of š„: š„ 36: log 36 log 36 35 2
š„ 1: log 1 log 1 35 2 X
Final solution: š± šš
Starting equation: log 5š„ 5 log 3š„ 7
Take 6 to the power of both sides: 6 6
Simplify: 5š„ 5 3š„ 7
Add 5: 5š„ 3š„ 12
Subtract 3š„: 2š„ 12
Divide by 2: š„ 6
Test the solution of š„: log 5 ā 6 5 log 3 ā 6 7
Final solution: š± š
These terms are both Invalid because negative
numbers are not in the domain of the log function.
Note: To test the
solutions you derive,
use the original
equation or a
simplified form of the
original equation.
PreCalculusChapter3PracticeTest Name:____________________________
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Starting equation: log š„ 5 1 log š„
Add log š„ to both sides: log š„ 5 log š„ 1
Combine log terms: log š„ 5 ā š„ 1
Take 14 to the power of both sides: 14 ā 14
Simplify: š„ 5 ā š„ 14
Multiply terms: š„ 5š„ 14
Subtract 14: š„ 5š„ 14 0
Factor: š„ 2 š„ 7 0
Determine solutions for š„: š„ 2, 7
Test the solutions of š„: š„ 2: log 2 5 log 2 1
š„ 7: log 7 5 log 7 1 2 X
Final solution: š± š
Starting equation: 4 8
Convert 4 and 8 to powers of 2: 2 2
Simplify: 2 2
Equate the exponents: 2š„ 18 3š„ 6
Subtract 2š„: 18 š„ 6
Add 6: 24 š„
Final solution: š± šš
These terms are both Invalid because negative
numbers are not in the domain of the log function.
Use this equation
to test your
solutions below.
Note: You can only
equate exponents
when the bases are
the same.
PreCalculusChapter3PracticeTest Name:____________________________
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For an exponential function of the form: š š„ š š, start with the parent function: š š„ š ,
and shift the function ā units right (shift left if ā is negative) and š units up (shift down if š is negative). The horizontal asymptote is at š¦ š. One way to do all of this is to:
Identify two points and any asymptotes on the parent function graph.
Shift the two points and any asymptotes to set up the desired graph.
Graph the desired function.
Note: in the graphs that follow, all shifts are shown with purple arrows.
The parent function is š¦ 4
Shift the parent function 2 units right and 2 units up.
Shift the horizontal asymptote of the parent, š¦ 0, 2 units up to become š¦ 2.
PreCalculusChapter3PracticeTest Name:____________________________
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The parent function is š¦ š
Shift the parent function 2 units right and 4 units up.
Shift the horizontal asymptote of the parent, š¦ 0, 4 units up to become š¦ 4.
The parent function is š¦ 0.6
There are no shifts.
Plot two points that are easy to identify on the curve: 0, 1 and 2, 0.36 .
Run a decreasing exponential curve (since 1) through the points.
PreCalculusChapter3PracticeTest Name:____________________________
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The parent function is š¦ log š„
Shift the parent function 2 units left. There is no vertical shift. š š„ log š„ 2 0
Shift the vertical asymptote of the parent, š„ 0, 2 units left to become š„ 2.
We are given: š $900 š 0.12 š 4 (quarterly compounding)
š” 6 years ā 4 quarters 24 periods
š“ š 1šš
ā900 ā 1
0.124
900 ā 1.03 $š, ššš. šš
We are given: š $4,000 š 0.07 continuous compounting (use Pert formula)
š” 5 years
š“ šš 4000 ā š ā . 4000 ā š . $š, ššš. šš
PreCalculusChapter3PracticeTest Name:____________________________
12 | P a g e
There are two steps to problems like this:
1) Find the value of š based on the change in temperature over the first 5 minutes.
2) Find the final temperature, š, after 19 minutes.
What are the variables?
š» is the temperature of the object (mostacolli) at any point in time. š» changes as š (time)
changes.
šŖ is the current temperature of the environment (e.g., the room).
š»š is the starting temperature of the object (mostacolli).
š is the rate of change in the temperature of the object (mostacolli) over time.
š is the time elapsed since the event (removal from the oven) occurred. Initially, š 0.
Step 1: Find the value of š.
We are given: š 375 š¶ 68 š” 5 š 327
Starting equation: š š¶ š š¶ ā š ā
Substitute known values: 327 68 375 68 ā š ā
Subtract 68 from both sides: 259 375 68 ā š ā
Simplify: 259 307 ā š
Divide by 307: 0.843648 š
Take natural logs: ln 0.843648 ln š
Simplify: 0.17002 5š
Divide by 5 to obtain š: š 0.0340039
Step 2: Find the final temperature, š», after 19 minutes.
We are given: š 375 š¶ 68 š” 19 š 0.0340039
Starting equation: š š¶ š š¶ ā š ā
Substitute known values: š 68 375 68 ā š . ā
Calculate the final value of š: š» šššĀ° š
PreCalculusChapter3PracticeTest Name:____________________________
13 | P a g e
There are two steps to problems like this:
1) Find the value of š based on the halfālife of 710 years. 2) Find how much would be left after 200 years.
What are the variables?
The formula for exponential decay is: š“ š“ š , where:
š“ is the amount of substance left at time š”. š“ is the starting amount of the substance.
š is the annual rate of decay. š” is the number of years.
Step 1: Determine the value of š
We are given: š” 710, (because we are given a āhalfālifeā)
Starting equation: š“ š“ š
Divide by š“ : š
Substitute in values: š
Take natural logs: ln 710 š
Divide by 710: š
0.00097626
Step 2: Find how much is left after 200 years
We are given: are given: š” 200, š“ 50 grams š 0.00097626
Starting equation: š“ š“ š
Substitute in values: š“ 50 ā š . ā šš. ššš grams
PreCalculusChapter3PracticeTest Name:____________________________
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This is a simple substitution problem. Substitute 9 for š” and calculate the solution.
š 987,000
1 1,449 ā š . ā
87,0001 1,449 ā š . šš, ššš
There is only one step to this problem:
1) Find the value of š based on the population change from 1984 to 1989.
What are the variables?
The formula for exponential decay is: š“ š“ š , where:
š“ is the population at time š”. š“ is the starting population.
š is the annual rate of growth. š” is the number of years.
Step 1: Determine the value of š
We are given: š“ 37, š“ 30, š” 5
Starting equation: š“ š“ š
Substitute in values: 37 30š ā
Divide by 30: š
Take natural logs: ln 5š
Divide by 5: š š. ššššššš
Round to 3 decimal places: š š. ššš
PreCalculusChapter3PracticeTest Name:____________________________
15 | P a g e
There are two steps to problems like this:
1) Find the value of š based on the halfālife of 5600 years. 2) Find how old the fossil is when there is 18% left.
What are the variables?
The formula for exponential decay is: š“ š“ š , where:
š“ is the amount of substance left at time š”. š“ is the starting amount of the substance.
š is the annual rate of decay. š” is the number of years.
Step 1: Determine the value of š.
We are given: š” 5600, (because we are given a āhalfālifeā)
Starting equation: š“ š“ š
Divide by š“ : š
Substitute in values: š
Take natural logs: ln 5600 š
Divide by 5600: š
0.000123776
Step 2: Find how old the fossil is when there is 18% left.
We are given: are given: 18% left, š 0.000123776
Starting equation: š“ š“ š
Divide by š“ : š
Substitute in values: 0.18 š . ā
Take natural logs: ln 0.18 0.000123776 ā š”
Divide by 0.000123776 : š .
.šš, ššš years old