1 Math 30+ Review #5 On the test, for full credit, you will need to show all your work, clearly circle or box your answers when applicable, and round final answers to three decimal places when...

1

Math 30+ Review #5

On the test, for full credit, you will need to show all your work, clearly circle or box your answers when applicable, and
ound final answers to three decimal places when approximating (unless there are fewer than 3, or unless 3 decimals is
inappropriate for the context of the problem). You will be allowed to use a pencil, eraser, ruler (or other straight edge),
and scientific calculator. The computers will not be required and, hence, not permitted for this test.

PS 10.1: Integer Exponents
1) Simplify each of the following expressions
(i) (7?2)2 = 72(?2)2 (ii)
8?3?4
2?−7
= 4?3?4?7 (iii)
(3??5)
2
(2?−2?2)3
=
32?2(?5)
2
23(?−2)3(?2)3

= 49?4 = 4?3?11 = 9?
2
?10
8?
−6
?6

= 9?
2
?
6
?10?−6
8

= 9?
8
?4
8
2) Let ?(?) = 4(5)?. Find the following.

(i) ?(−2) = 4(5)−2 = 4
5
2 =
4
25
(ii) ?(0) = 4(5)0 = 4 ∙ 1 = 4


(iii) ?(5) = 4(5)5 = 4 ∙ 3125 = 12,500


PS 10.3: Graphing Exponential Models
3) Which of the following graphs belongs to ?(?) = 4(5)??
If we assume that one of these graphs belong to the function ?(?) = 4(5)? then, amongst other
considerations, the graph must have the appropriate ?-intercept, which is (0,4): ?(0) = 4 (see (ii) above).
This leaves graph b.

(a) (b) (c)


(Note: This solution depends on the assumption that at least one of these graphs belongs to ?(?) = 4(5)?. I
say this because it is possible for some a
itrary, exponential looking graph to pass through (0,a) and still not
e the graph of ?(?) = ???).
2

PS 10.5: Exponential regression Model
4) Refer to the association given below to answer the questions that follow.
(i) Describe the 4 characteristics of the association. (Don’t forget to consider the co
elation coefficient.
Outliers: Hard to determine – let’s assume none until we know what shape is going to be.
Shape: Nonlinear for sure – possibly exponential
Strength: Visually, it looks moderately strong as an exponential association. The exponential co
elation
coefficient is ? = XXXXXXXXXXwhich suggests it’s actually quite strong as an exponential association.
Direction: it’s a positive association
(ii) Produce the regression equation with the coefficient and base rounded to three decimal places.
An exponential regression equation is of the form �̂� = ???. In this case, ? = XXXXXXXXXXand ? = XXXXXXXXXX.
So, the regression equation, with ? and ? rounded as desired, is �̂� = XXXXXXXXXX)?
(iii) By the way, what is so special about the regression model anyways?
For a given association shape, it is the model that does the best job describing the association. We didn’t talk
about this in this class, but the fact that the regression model is the model that has the smallest sum of squared
esiduals is what allows it to do the best job of describing the association.
(iv) Use this model to predict the response value when the explanatory value is 3. Round to three decimal places.
�̂� = XXXXXXXXXX) ≈ 3.833
(v) What is the residual for the prediction you made in part (iv)?
Residual = ? − �̂� = 3.795 − 3.833 = −0.038

IA 2.3: Logarithmic Functions
5: Let ?(?) = log4(?).
(i) Find both ?(64) and ? (
1
16
).

?(64) = log4(64) = 3, ??????? 4
3 = 64.

? (
1
16
) = log4 (
1
16
) = −2, ??????? 4−2 =
1
42
=
1
16
.








3

(ii) Which of the following graphs belongs to ?? How do you know?
(a) (b) (c)


Graph (c) is the graph of ?(?) = log4(?). The reason we know this is because, for any base ?,
log?(?) = 1, ??????? ?
1 = ?. This means that, for every function of the form ℎ(?) = log?(?) , (?, 1) is a point
on its graph. So the graph of ?(?) = log4(?) is the one that goes through the point (4, 1).
One word of caution: This is a contrived exercise where the assumption is made that at least one of these
graphs is the graph of ?(?) = log4(?), and that we just have to find which of these it is. It is possible in general,
however, that none of these would be the graph, as there are other functions whose graphs look logarithmic
and pass through the point (4, 1), while not actually being the function ?(?) = log4(?).

IA 2.4: Properties of Logarithms
6: Solve: 2 + log4 (
?
2
) = XXXXXXXXXX: Solve for ?: log?(81) = 4
2 + log4 (
?
2
) = XXXXXXXXXXlog?(81) = 4
log4 (
?
2
) = XXXXXXXXXX = ?4
?
2
= 42 = XXXXXXXXXXboth (−3)4 = 81 ??? 34 = 81, but only
? = 2 ∙ 16 = XXXXXXXXXXpositive numbers are acceptable as bases for
XXXXXXXXXXLogarithms. So ? = 3.

8: Solve. Round your approximation to four decimal places. Then check that your solution approximately satisfies the
original equation: XXXXXXXXXX)? = 101.27
Note: Don’t approximate anything until the very last step where you throw everything into a calculator.
XXXXXXXXXX)? = 101.27
4.76? =
101.27
5.91

log(4.76?) = log (
101.27
5.91
)
? log(4.76) = log (
101.27
5.91
)
? =
log (
101.27
5.91
)
log(4.76)
≈ 1.8210






4

IA 2.5: Using the Power Property with Exponential Models to Make Predictions
9: How long will it take a group of 24 wild ra
its, which increases in number by 19% every month, to reach a
population size of 2,000,000? Approximate your answer to one decimal place.
(Hint: look at example 1 from the IA 2.5 notes)
Since the group size increases by the same percentage, 19%, each month, the growth is exponential. So if we set
?(?) to be the population size at time ?, in months, then ?(?) = ??? for some coefficient ? and base ?.
Since the beginning population is 24, we know 24 = ?(0) = ??0 = ? ∙ 1 = ?. That is, ? = 24.
Since the population increases in size by 19% each month, we know the base is ? = XXXXXXXXXX = 1.19.
So ?(?) = XXXXXXXXXX)?
Now we just need to find ? such that ?(?) = 2,000,000:
24(1.19)? = 2,000,000
1.19? =
2,000,000
24

log(1.19?) = log (
2,000,000
24
)
? log(1.19) = log (
2,000,000
24
)
? =
log (
2,000,000
24 )
log(1.19)
≈ 65.14
It will take about 65 months for the original group of 24 ra
its to multiply to a population size of 2,000,000.
(Note: it was unclear when I looked at the source for this data whether or not the 19% growth rate accounts for
environmental pressures like getting run over by cars or eaten by coyotes.)

10: Physicians use gallium citrate-67 to detect certain types of cancer. Gallium citrate-67 has a half-life of 3.25 days –
some is lost to radioactive decay, and some is removed through the digestive and urinary tracts. A patient who is
east feeding is injected with the radioactive substance.
a) let ?(?) be the percentage of gallium citrate-67 that remains in the patient’s body at ? days since she was
injected. Find an equation of ?.
When a substance decays with a measurable half-life, the underlying reality is that the decay is exponential, so
?(?) = ??? for some numbers ? and ?. Since it is exponential decay, we’re expecting ? to be less than 1. As
we’re timing how long the substance has been in the body and measuring what percentage of it remains, at the
precise moment of injection time equals zero and 100% of the substance remains:
100 = ?(0) = ??0 = ? ∙ 1 = ?
So ? = 100.
Since the half-life is 3.25 days, meaning 50% of the substances remains after 3.25 days,
?(3.25) = 100?3.25 = 50
?3.25 = 0.5
Since raising ? to the 3.25th power yields 0.5, ? is the 3.25-th root of 0.5. So
? = √0.5
3.25

Try not to stress out too much about what it means to be a 3.25th root of something – we never really talked
about this kinda stuff in any meaningful way in class. It suffices to say there is a button on your calculator that
will calculate the 3.25th root of 0.5 for you. A typical scientific calculator does this when you enter the following
sequence:
0.5  √?
?  3.25  =
(you may have to use the 2nd button to activate the √?
? feature: 0.5  2nd  √?
?  3.25  =)
5

Doing this in my calculator yields ? = √0.5
3.25
≈ XXXXXXXXXX

I rounded to 6 decimal places because,