Test 3 Review
Miscellaneous Algebra Notes:
• Constant factors do not mix with the base: 4(2x ) 6= 8x . (It’s 22 · 2x instead.)
• Negative exponents can be written as reciprocals, producing denominators in equations.
Sample: ex + e−x = 2 is the same as ex + 1/ex = 2; multiply ex to get e2x + 1 = 2ex .
• Exponential and log laws can be used in two different ways:
– To split up complex exponentials or logs, like 32x+y = 32x 3y = (3x )2 3y
– To combine into one exponential or logarithm, like 2 log(x) − log(y) + 4 log(z) = log(x2 z 4 /y).
When does it help to split exponentials? When does it help to combine?
– The laws for exponentials and logs are opposites. For example, ex+y = ex ey splits a sum into a
product, but log(xy) = log(x) + log(y) splits a product into a sum.
• With equation-solving or with domain questions, make sure any number you plug into a log is positive.
• If you can’t easily cancel one base, use any log with the power property. (See Section 4.6.)
Sample: Solve 2x = 7x−1 with either log or ln; get x log(2) = (x − 1) log(7) or x ln(2) = (x − 1) ln(7).
Summaries to keep in mind:
• We use nth roots to cancel exponents, but we use logs to cancel bases.
• Combine exponentials or logs with the same base, especially to solve equations.
• Some equations feature a quadratic in something complicated. A substitution can help break the problem
into two parts.
Sample: In 22x − 5(2x ) + 4 = 0, we substitute u = 2x to get u2 − 5u + 4 = 0. Find the solutions for u
(u = 1, 4). Finally, go back to u = 2x to find x: 2x = 1, 4, so x = log2 (1), log2 (4).
• The formula A = A0 ert (or y = y0 ekx , etc.) is very common with exponentials.
– In many scenarios, the initial value y0 cancels.
– Commonly, one phase finds the rate constant r (or k) first from a data point (like half-life).
– Some problems make you write two exponential formulas equal to each other (using different rates
and/or initial values). Divide one exponential to get the exponentials on the same side.
Selected WebQuiz 5 and 6 Problems with Hints
For an overview of more problems, visit my website and look for an extra downloadable handout.
WebQuiz 5
#2: “Find the zeros of f . (Enter your answers as a comma-separated list.)”
f (x) = x2 e7x + 8xe7x + 20e7x + xe7x
We want to solve f (x) = 0. Since every term has an e7x factor, pull it out! You get e7x times a quadratic
expression; set each equal to 0. But note that there are no zeroes of e7x ; why not?
#3: “Find an exponential function of the form f (x) = ba−x + c that has the given horizontal asymptote
and y-intercept and passes through point P : y = 8; y-intercept 48; P (3, 13)”
The asymptote tells us c = 8. You plug in the y-intercept with x = 0: 48 = ba−0 + 8 and find b. Lastly, plug
in the point P to find a. (The solution uses a cube root.)
#4: “Find the inverse of the one-to-one function f (x) = −e−x + 4 and find the domain of f −1 (x).”
To find inverses, we write y = −e−x + 4 and solve for x in terms of y. Don’t try to do too many steps at
once here! Cancel only one operation at a time. Note that the two minuses in −e−x are dealt with in different
steps of the solution; you’ll have to divide that outer − before you can cancel the e base.
#10 version 1 (modified): “Determine the effective interest rate if a nominal rate of 5% is compounded
weekly. (Assume 52 weeks in a year.) Express your answer as a percent.”
#10 version 2: “Determine the per annum interest rate required for an investment with continuous
compound interest to yield an effective rate of 5%. Express your answer as a percent”
In these problems, we get the same amount A from two different interest formulas. In both versions,
“effective” rate uses a simple interest formula compounded once, i.e. A = P (1 + r)t . However, version 1 also
uses a weekly compounding formula (i.e. n = 52), and version 2 uses the continuous compounding formula. All
of these formulas will use the same P and t, and those values should cancel in the equations!
Read carefully to figure out which formula has the known rate and which has the unknown rate.
WebQuiz 6
#1: “Mark each of the following logarithm ’properties’ as True, if true (for all real numbers x and y for
which the expressions are defined) or False, if not.”
(a) ln(x/y) = ln(x)/ ln(y)
(b) ln(0) = 1
(c) ln(x + y) = ln(x) ln(y)
(d) ln(ex ) = x
These problems are trying to see how well you recognize the parts of the algebra laws. For instance, part (a)
cannot be right, because the log laws never use the same operation before and after splitting the logs. In fact,
when you study these, ask yourself: can you fix the false statements with small changes?
p
#2: “Evaluate loga ( 3 ax8 /y 9 ) given that loga (x) = 3 and loga (y) = −7.2.”
This problem is giving you the values of simple logs. Thus, we should split up our complex log into simple
logs! First, I’d use the power law with the cube root (it’s a 1/3 power). Next, split up the products and
quotients, and lastly, use the power law again.
#4 (two versions): “Solve these equations for x.”
(a) a− loga (x) = 0.2
(b) ln(x + 3) = ln(x) + log4 (2)
The first one has base a and logarithm with base a, so these should cancel! The − in the exponent stays,
though: you get (x)−1 = 0.2.
For the second one, let’s bring our logs together with the unknown x: ln(x + 3) − ln(x) combines to ln((x +
3)/x). You can simplify log4 (2)... think of roots. After canceling the ln, we get a rational equation with
(x + 3)/x; start by multiplying x through the equation.
Don’t forget that with log equations, you should double-check to make sure your answer is actually legit!
Plug it in to the logs to make sure their insides are still positive.
#6: “Use common logarithms to solve for x in terms of y.”
y=
102x − 10−2x
3
First, we clear out denominators: multiply 3 through, then multiply 102x (since 10−2x has 102x as denominator). You get 104x − 1 = (3y)102x . Since we have both 102x and its square 104x , this is quadratic in u = 102x .
Solve for u using the Quadratic Formula, then replace u back with 102x and finish solving for x.
#7: “If a radioactive isotope has half-life 1599 years, and 1.7 grams remain after 1000 years, what was its
initial amount?”
We use an exponential decay equation A = A0 ekt , and we want to know A0 . First, though, we find the rate
constant k by using the half-life: A = A0 /2 when t = 1599. Now that we know k, we can put in the other point
we’re given, and you can solve for A0 .
#9: “In 1944, the Gutenberg-Richter Law established the equation: N = 10a−M . Here, a is a constant
and M represents a Richter scale magnitude, while N represents the number of earthquakes in one region of
magnitude at least M .
(a) Compute a, assuming that there are 2 earthquakes of magnitude at least 6 per year.
(b) Using the assumption from part (a), how many earthquakes of magnitude at least 4 will occur per year
in that region?”
Fortunately, you’re given a formula to use! We have to be careful with what each variable does. N is the
number of earthquakes, and M is magnitude. Thus, part (a) says N = 2 and M = 6; plug these in and solve
for a. For part (b), they’re giving us M = 4, and we already know a, so we can get N .