Midterm 3 Practice Problems
The following problems are a bit more difficult than what I’d expect on the exam. If you can complete
each of them, I’d expect that you’ll be in good shape for the exam.
Solutions can be found at the end of the document. One of each sort of problem has a complete solution;
the rest just have answers. It is important to note that just giving an answer will not be acceptable on the
exam; you must show your work in clear detail.
1. Simplify the following:
√
7
log12
√
3
144x9
!7/10
12x−2
2. Simplify the following:
√
11
log23
√
4
23x2
!4/13
529x1/3
3. In the first few weeks of the zombie apocalypse, scientists determine that the rate of infection is
proportional to the number of infected people. The outbreak began with just one person; two weeks later,
10, 000 people were infected. Assuming this rate of growth continues, how long will it take to infect all
7, 000, 000, 000 people on Earth?
4. A certain radioactive isotope is kept in a vial. After 1000 days, only one-third of the original amount
is left. Find the half-life of this isotope.
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5. log2 (2x − 1) + log2 (x2 − 9) = log2 (x + 3) + log2 (x2 + 1) + 1. Solve for x.
6. log3 x − log3 (x + 1) = log3 (x − 1) − 1. Solve for x.
2
x
− 2).
7. Find the domain of f (x) = ln( x+3
8. I place 600 grams of a radioactive isotope with a half-life of 10 months in a box. At the same time,
I put $100 in a bank account. If interest in the bank account is compounded continuously, what should the
interest rate be so that the number of grams of isotope left and the number of dollars in the bank account
are exactly equal after 2 years?
(This problem is very difficult. Take it one step at a time, don’t worry about jumping straight to the
answer.)
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Solutions
Solution 1. Take this one step at a time. The goal is to make the thing inside the logarithm as simple
as possible; so we should try to get rid of as much as we can, one operation at a time. The biggest operation
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power, so let’s get rid of that. Pulling that out of the logarithm turns exponentiation into
there is that 10
multiplication - so we’re left with:
!
√
7
144x9
7
log12 √
3
10
12x−2
The next operation is that division. Pulling division out of the logarithm turns it into subtraction; so
now we have:
√
√
7 7
3
log12 144x9 − log12 12x−2
10
Let’s concentrate on that leftmost one first. No reason, just because I like going left-to-right. That
radical is the biggest problem now. We don’t have any rules for roots, but we do have rules for powers - so
let’s re-write it with a power.
√
7 3
log12 (144x9 )1/7 − log12 12x−2
10
Now we can pull out that 1/7th power:
√
7 1
3
log12 (144x9 ) − log12 12x−2
10 7
It’s tempting to pull out that 9th power, but the 144 is in the way! So we need to pull that out first.
Logarithms turn multiplication into addition, so now we have:
√
7 1
3
9
−2
log12 (144) + log12 (x ) − log12 12x
10 7
Now we have three logs. Again, let’s concentrate on the left one, log12 (144). The only way this is going
to simplify is if 144 happens to be an easy power of 12. Let’s try it: 120 = 1, 121 = 12, 122 = 144 - there it
is. So log12 (144) = 2. Now we have
√
7 1
3
9
−2
2 + log12 (x ) − log12 12x
10 7
Now let’s look at the next logarithm, log12 (x9 ). Now we can pull out that exponent, leaving:
√
7 1
3
−2
(2 + 9 log12 x) − log12 12x
10 7
log12 x isn’t going to simplify any √
further - that’s as simple as we could hope for. So let’s turn our
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attention to that last logarithm, log12 12x−2 . Again, roots are no good, but exponents are easy - so let’s
turn that into log12 (12x−2 )1/3 . Now we can pull out the exponent, leaving:
1
7 1
(2 + 9 log12 x) − log12 (12x−2 )
10 7
3
Now, the biggest operation inside that logarithm is multiplication - let’s pull out that 12.
7 1
1
−2
(2 + 9 log12 x) −
log12 12 + log12 (x )
10 7
3
3
We have two logarithms now. Once again, let’s concentrate on the leftmost one, log12 12. Again, the
only way this is going to simplify is if 12 happens to be a nice power of 12 - and oh, hey, 12 = 121 . So
log12 12 = 1. Now we have:
1
7 1
−2
(2 + 9 log12 x) −
1 + log12 (x )
10 7
3
One log left. All that’s in there is an exponent, so let’s pull that out.
7 1
1
(2 + 9 log12 x) − (1 − 2 log12 x)
10 7
3
Now, all the logs are as simple as they get. Now it’s just a matter of getting rid of all of those pesky
parentheses. Push the 17 and 13 in:
7 2 9
1 2
+ log12 x − + log12 x
10 7 7
3 3
Careful with your signs here! Next, let’s consolidate the stuff inside. We can combine the
and we can combine the log12 x terms.
1
41
7
− +
log12 x
10
21 21
Finally, let’s bring in the
7
10 .
−
1
41
+
log12 x
30 30
4
2
7
and the 31 ,
18
Solution 2. The final solution is − 143
+
13
429
log23 (x).
Solution 3. This is a word problem, so the first step is to translate into math. Looking at that first
sentence, the phrase “the rate of infection is proportional to the number of infected people” should make
you think “oh, hey, exponential growth!”. So we’re using exponential growth here, the formula for which is
P (t) = P0 ekt
What do all those pieces mean for us? P (t) means “population at time t”, but population of what? Well,
the population that’s growing is the population of zombies, so that’s what P (t) is: the number of zombies
at time t.
P0 is the initial population, or the number of zombies at time 0.
k is the rate of growth.
t is the time since the first infection; we need to pick units for this. I’ll pick days; weeks would also make
plenty of sense.
The next sentence says that the outbreak started with just one zombie: that is, the population at time
0 is 1. So P0 = 1.
The third sentence says that there were 10, 000 infected two weeks later. That’s t = 14 days (this is
where our choice of units for t is important). So P (14) = 10, 000. But P (t) = 1 · ekt = ekt , so 10, 000 = e14k .
Finally, the question is, when is the number of zombies equal to 7, 000, 000, 000? In math, that’s “for
what t is P (t) = 7, 000, 000, 000?” In other words, “for what t is ekt = 7, 000, 000, 000?” In order to find
that out, we’re going to want to solve for t - but for that, we’ll need to know k.
We can get k from our earlier equation, 10, 000 = e14k . Take ln of both sides, giving ln(10, 000) = 14k.
1
ln(10, 000). There are two important things to notice here: First, it doesn’t matter whether I
So k = 14
simplify k or anything - k isn’t my final answer, so I can write it however I want. On the other hand, I
shouldn’t plug this into a calculator - if I do, I’ll be rounding, and that error will get amplified later. If I’m
going to evaluate at all, I need to wait until the very last step.
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Now I know k, so now I know that ekt = e 14 ln(10,000)t . So my second equation becomes:
1
e 14 ln(10,000)t = 7, 000, 000, 000
Solving:
1
ln(10, 000)t = ln(7, 000, 000, 000)
14
t=
And so our final answer is:
days.
ln(7,000,000,000)
1
14 ln(10,000)
ln(7, 000, 000, 000)
1
14 ln(10, 000)
days after initial infection. If you’re curious, that’s about 34.5
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Solution 4. The word “radioactive” should make you think “radioactive decay equation”. So we’ll be
using N (t) = N0 e−kt . Again, let’s name the variables.
• N (t) is the amount of isotope left at time t.
• N0 is the amount of isotope at time 0.
• k is the rate of decay.
• t is the time in days since the beginning of the experiment. Other choices of unit and starting point
are possible, but this seems like the best one.
Now, we know that after 1000 days, 1/3 of the original amount is left. Using the definitions we just gave,
“1/3 of the original amount” is “1/3 of N0 ”, which is 13 N0 . So N (1000) = 31 N0 . Therefore:
1
N0 = N0 e−1000k
3
Now, we don’t know N0 , so we have to hope it won’t matter - and sure enough, we can cancel it from
both sides.
1
= e−1000k
3
1
ln( 31 ). So we now have:
Solving for k, we get k = − 1000
1
1
N (t) = N0 e 1000 ln( 3 )t
Now, let’s turn our attention to the question. The question asked “what is the half-life?” That’s the
same as asking “How long does it take for there to be only half of the original left?” In other words, “How
long does it take for there to be only 12 N0 left?” Or, “For what t is N (t) = 21 N0 ?”
1
1
1
1
N0 e 1000 ln( 3 )t =
1
N0
2
Again, we can cancel N0 .
e 1000 ln( 3 )t =
1
2
Solving for t, we get:
t=
After some clever simplification, I get: t =
ln( 12
1
1
1000 ln( 3 )
1000 ln 2
ln 3 .
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So my final answer is: The half-life is
1000 ln 2
ln 3
days.
Solution 5. As always, take this one step at a time. Looking at this equation, what I see is lots of logs
on both sides; that makes me think about things that look like “log(a) = log(b)”. If we had that, we could
strip away the logs and get “a = b”. So let’s try for that - to do that, we need to have only one log on each
side of the equation. Remember, addition of logarithms is the same as multiplying inside logarithms, so we
have:
log2 ((2x − 1)(x2 − 9)) = log2 ((x + 3)(x2 + 1)) + 1
We need to deal with that 1 somehow. If it had a log in front of it, we could just use our log properties
again - so let’s turn it into a logarithm. We want a number a so that 1 = log2 (a). Remember, that’s the
same as saying 21 = a. So a = 2. In other words, 1 = log2 (2). So we can write:
log2 ((2x − 1)(x2 − 9)) = log2 ((x + 3)(x2 + 1)) + log2 2
Using our product rule again, we have:
log2 ((2x − 1)(x2 − 9)) = log2 (2(x + 3)(x2 + 1))
There! Now we have log(something) = log(something else), so we can strip away the logs:
(2x − 1)(x2 − 9) = 2(x + 3)(x2 + 1)
Now all the logs are gone; there are no exponentials; there’s not even any division going on. That means
this is a polynomial problem now. The best way to solve a polynomial problem, usually, is to expand and
sort everything over to one side, so that we can use the Zero Product Property. Expanding, we get:
2x3 − x2 − 18x + 9 = 2x3 + 6x2 + 2x + 6
Moving everything to the right and combining like terms:
0 = 7x2 + 20x − 3
(Why use the right, and not the left? No reason, it just looks a little prettier this way.) Now we have
a quadratic equation. We have three techniques we can use to solve this: factoring, completing the square,
and the quadratic formula. In this case, the quadratic formula seems appropriate. That gives us x = −3, 71 .
We’re not done yet. Logarithms have tricky domain problems, so we need to go back and check
whether these values of x give sensible results in the original equation.
If we plug in x = −3, that “log2 (x + 3)” involves taking a log of 0, which is illegal. So −3 is not a
solution.
If we plug in x = 17 , that “log2 (2x − 1)” is really “log2 (− 57 )”, which is a log of a negative number; also
illegal. So 17 is not a solution.
To summarize: We found that the only possible solutions were −3 and 71 . We also found that neither of
them were actual solutions. Therefore, the equation does not have a solution.
Note: On the exam, it is reasonable to expect that any problem you are asked to solve will have a solution;
in that respect, this problem is more difficult than anything you would see on the exam.
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Solution 6. x =
x−1
x
x+1 = 3 .
√
3+ 13
.
2
If you solve it the way I did, you should find yourself solving the equation
Solution 7. Well, first, the function is clearly undefined when x + 3 = 0, because then we’re dividing
by zero; so x = −3 can’t be in our domain.
The more important piece, though, is this: ln is only defined for positive values. So we need the thing
inside the ln to be positive. In other words, we need:
x2
−2>0
x+3
This is a rational inequality, so the strategy for solving it begins with finding boundary points. First,
note that the inequality is undefined when x + 3 = 0, so x = −3 is a boundary point.
Next, we need to look for cases when the inequality is actually an equality. So we’re trying to solve:
x2
−2=0
x+3
With some rearranging, we get the following:
x2 = 2x + 6
√
Moving everything to the left, we have x2 − 2x − 6 = 0. Apply the quadratic formula to obtain x = 1 ± 7
(I’m skipping a lot of arithmetic here, of course).
√
√
So we have three boundary points: −3, 1− 7, and 1+ 7. That divides the number line into four pieces.
−4 is a suitable testpoint
for the leftmost piece,
4 for the rightmost. The middle two are a bit harder;√but
√
√
remember, 7 < 9, so 7 < 3. Therefore √
1 − 7 > −2,
√ which makes −2 suitable for the interval (−3, 1 − 7).
And 1 falls right smack in between 1 − 7 and 1 + 7.
√
√
Plugging each of these into the inequality results in
√ (−3, 1 − 7) ∪ (1 + 7, ∞); both intervals are open,
because −3 makes the inequality undefined, while 1± 7 both
√ make the
√ inequality equal to zero, not greater.
−3 is already excluded, so our final answer is (−3, 1 − 7) ∪ (1 + 7, ∞).
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Solution 8. The heart of this problem is “how much isotope is left” and “how much money is in the
account”. Let’s answer each one independently.
How much isotope is left? This is radioactive decay, so we’ll be using the equation N (t) = N0 e−kt . N
means “number of grams remaining at time t”. N0 means “the number of grams at time 0”, and t means
“time in months since the beginning of the problem”. I could have chosen years, or days; months makes the
most sense to me because of that “10 months” floating around.
N0 is 600, because the problem says so. Which means the only unknown left is k. Let’s use the other
piece of information we have to find k: the half-life. “The half life is 10 months” means “when t is 10, half
of the original amount is left”. In other words, N (10) = 21 N0 . So 12 N0 = N0 e−10k . Cancelling N0 from both
1
ln 2. So we now have:
sides and solving for k, we get k = 10
1
N (t) = 600e− 10 ln 2·t
6
At the 2-year mark, t is 24, so there are 600e 5 ln 2 grams remaining after 2 years.
Now, how much money is in the account? This is continuous compound interest, so we need to use the
equation A(t) = P ert . A(t) is the amount at time t. P is the initial amount, r is the rate, and t is the time
in years since the beginning. Here I’m using years, because I want the annual interest rate, and any other
unit will be very inconvenient.
P is 100, so we have A(t) = 100ert . We’d like r, but that’s the goal of the problem - so we’ll hold off on
that. In the meantime, we know we want A(2). A(2) = 100e2r .
Finally, we want these two numbers to be the same. That is,
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100e2r = 600e 5 ln 2
There’s only one variable here, r - and that happens to be the one we want! So let’s solve for that. That
gives us r = 12 ln 6 − 65 ln 2.
We’re almost done - we’d just like this as a percentage. That’s just 100 times the rate, so our answer is
100( 12 ln 6 − 56 ln 2, or 50 ln 6 − 120 ln 2.
Finally, if this is the computer exam, we should plug this into a calculator. That gives us about 6.4
percent.
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