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Chapter Review · a rapid pre-test review (speedrun)

Polynomials, Exponentials, and Logarithms: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Polynomial
A finite sum of terms, each a real coefficient times a nonnegative-integer power of xx. No variable in a denominator or under a root, and no negative or fractional exponent.
Term and coefficient
A term is one coefficient times one power of xx; the coefficient is its numerical factor with its sign, as 5-5 is in 5x2-5x^2.
Degree
Of a term, its power of xx (7x7x is degree 11, a constant degree 00); of a polynomial, the largest degree among its terms.
Leading term, leading coefficient, constant term
The highest-degree term, its coefficient (never 00), and the x0x^0 term: in 6x3x5+46x^3 - x^5 + 4, x5-x^5, 1-1, and 44.
Standard form
Terms in descending order of degree, highest power first and the constant last.
Monomial, binomial, trinomial
Polynomials of one, two, and three terms.
Like terms
Terms carrying the same power of xx. Only these merge into one term.
Closure
Adding, subtracting, or multiplying polynomials always gives another polynomial.
Exponential function
The variable in the exponent over a fixed base b>0b > 0 with b1b \ne 1: f(x)=abxf(x) = a \cdot b^{x}.
Power function
Variable in the base under a fixed exponent, as in x2x^2. The opposite arrangement, not exponential.
Principal PP, rate rr, frequency nn
The amount deposited, the yearly rate as a decimal (5%5\% is r=0.05r = 0.05), and the compounding periods per year.
The number ee
The irrational constant e2.71828e \approx 2.71828 that (1+1m)m\left(1 + \tfrac{1}{m}\right)^{m} climbs toward as mm grows.
Logarithm
An exponent. Read logb(y)\log_{b}(y) as "the exponent that turns bb into yy": bx=yb^{x} = y is exponential form, logb(y)=x\log_{b}(y) = x logarithmic form.
Common log and natural log
log(y)\log(y) with no base written means log10(y)\log_{10}(y); ln(y)\ln(y) means loge(y)\log_{e}(y).

Formulas and theorems

  • Combining like terms

    c1xk+c2xk=(c1+c2)xkc_1 x^{k} + c_2 x^{k} = (c_1 + c_2)x^{k}

    Use when The same power kk on both. Add the coefficients and keep the power; different powers never merge.

  • Subtracting a polynomial

    (A+B+C)=ABC-(A + B + C) = -A - B - C

    Use when A factor of 1-1 on EVERY term inside, however many there are.

  • Degree of a sum or difference

    deg(p±q)max(degp,degq)\deg(p \pm q) \le \max(\deg p, \deg q)

    Use when Equality unless pp and qq share a degree AND their leading coefficients cancel, the only way a degree can drop; then it falls to the next surviving power, or the result is 00, whose degree is left undefined.

    e.g. (5x3+x)(5x32)=x+2(5x^3 + x) - (5x^3 - 2) = x + 2: degree 11 from two degree-33 polynomials.

  • One term times one term

    (axm)(cxn)=acxm+n(a x^{m})(c x^{n}) = ac\,x^{m+n}

    Use when Multiply the coefficients, ADD the exponents. A negative sign folds into the coefficient.

    e.g. (4x5)(3x2)=12x7(-4x^5)(3x^2) = -12x^7.

  • Two binomials (FOIL)

    (a+b)(c+d)=ac+ad+bc+bd(a + b)(c + d) = ac + ad + bc + bd

    Use when Four products (First, Outer, Inner, Last), and only for two terms times two terms. The rule that always works: every term of the first times every term of the second, mnmn products in all. A lone term distributes the same way.

    e.g. (x5)(x+2)=x23x10(x - 5)(x + 2) = x^2 - 3x - 10.

  • Degree of a product

    deg(pq)=degp+degq\deg(pq) = \deg p + \deg q

    Use when Neither factor the zero polynomial. No exception: leading terms multiply, and nonzero times nonzero cannot cancel.

  • Exponential y=abxy = a\,b^{x}, and its graph

    y=abxy = a\,b^{x}
    Growth and decay share the point (0, a) and the asymptote y = 0A highlighted horizontal line low in the picture is labelled asymptote y equals 0. Two curves meet at a dot on a dashed vertical axis above it, labelled (0, a). The curve labelled growth arrives almost flat from the left, just above the highlighted line, then rises steeply on the right. The curve labelled decay is its mirror image: steep on the left, then almost flat just above the highlighted line on the right. Neither curve ever meets the highlighted line.(0, a)decaygrowthasymptote y = 0
    Text description

    Two exponential curves meeting at the point (0, a): the growth curve flattens toward the asymptote y = 0 on the left and climbs on the right, and the decay curve is its mirror image.

    Use when a>0a > 0, b>0b > 0, b1b \ne 1: a negative base breaks at fractional exponents, b=1b = 1 is a constant. Growth when b>1b > 1, decay when 0<b<10 < b < 1. Through (0,a)(0, a), horizontal asymptote y=0y = 0 (flat side left for growth, right for decay), domain all real numbers, range y>0y > 0, one-to-one, which is why bxb^{x} has an inverse, the logarithm.

    e.g. y=32xy = 3 \cdot 2^{x} grows from 33; y=5(13)xy = 5\left(\tfrac{1}{3}\right)^{x} decays from 55.

  • Adding inputs multiplies outputs

    bx+y=bxbyb^{x + y} = b^{x} \cdot b^{y}

    Use when b>0b > 0, any real xx and yy. With y=1y = 1: each unit step right multiplies the output by bb, the signature of an exponential.

    e.g. 3234=36=7293^{2} \cdot 3^{4} = 3^{6} = 729.

  • Simple interest

    A=P(1+rt)A = P(1 + rt)

    Use when rr a decimal, tt in years. Linear: the same PrPr is added every year, never interest on interest.

    e.g. 10001000 dollars at 5%5\% for 33 years: 1000(1+0.053)=11501000(1 + 0.05 \cdot 3) = 1150 dollars.

  • Compound interest

    A=P(1+rn)ntP=A(1+rn)nt\begin{gathered} A = P\left(1 + \frac{r}{n}\right)^{nt} \\ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} \end{gathered}

    Use when nn is compounding periods per YEAR (22 semiannual, 44 quarterly, 1212 monthly), rr a decimal, tt in years. The rate is divided by nn AND the exponent multiplied by nn. Annual is n=1n = 1: A=P(1+r)tA = P(1 + r)^{t}. Solving for PP needs no logarithm: the unknown is a plain factor, so division finishes it.

    e.g. 15001500 dollars at 6%6\% semiannually for 22 years: 1500(1.03)4=1688.261500(1.03)^{4} = 1688.26 dollars.

  • Continuous compounding

    A=PertA = P e^{rt}

    Use when The limit as nn grows without bound, and a CEILING: with r>0r > 0, compounding more often earns a little more each time, by shrinking amounts, but no finite schedule reaches PertPe^{rt}.

    e.g. 20002000 dollars at 3%3\% for 11 year: 2000e0.03=2060.912000e^{0.03} = 2060.91 dollars.

  • Definition of the logarithm

    logb(y)=x    bx=y\log_{b}(y) = x \iff b^{x} = y

    Use when Base b>0b > 0 with b1b \ne 1, input y>0y > 0; logb(0)\log_{b}(0) and the log of a negative are undefined, a positive base to a real power being always positive. Read off logb(1)=0\log_{b}(1) = 0, logb(b)=1\log_{b}(b) = 1.

  • The logarithm and the exponential undo each other

    blogb(y)=y,logb(bx)=xb^{\log_{b}(y)} = y, \qquad \log_{b}(b^{x}) = x

    Use when b>0b > 0 with b1b \ne 1. The first needs y>0y > 0; the second holds for every real xx.

  • The three laws of logarithms

    logb(MN)=logbM+logbNlogb ⁣(MN)=logbMlogbNlogb(Mk)=klogbM\begin{gathered} \log_{b}(MN) = \log_{b} M + \log_{b} N \\ \log_{b}\!\left(\frac{M}{N}\right) = \log_{b} M - \log_{b} N \\ \log_{b}(M^{k}) = k\log_{b} M \end{gathered}

    Use when One base throughout, b>0b > 0, b1b \ne 1, M>0M > 0, N>0N > 0, kk any real. Products become sums, quotients differences, powers coefficients. The power law moves an exponent on the INPUT out front, so logb(Mk)\log_{b}(M^{k}) is not (logbM)k\left(\log_{b} M\right)^{k}. There is NO law for the log of a sum.

  • Change of base, and solving for an exponent

    logb(y)=log(y)log(b)\log_{b}(y) = \frac{\log(y)}{\log(b)}

    Use when y>0y > 0, b>0b > 0, b1b \ne 1. One base top and bottom, ln\ln as readily as log\log. The answer is a QUOTIENT.

    e.g. log2(10)=log10log2=10.30103.32\log_{2}(10) = \frac{\log 10}{\log 2} = \frac{1}{0.3010} \approx 3.32.

  • Graph of y=logb(x)y = \log_{b}(x)

    The reflection of y=bxy = b^{x} across y=xy = x: through (1,0)(1, 0) and (b,1)(b, 1), vertical asymptote x=0x = 0, domain x>0x > 0, range all real numbers.

    The logarithm is the exponential reflected across the line y = xTwo curves on one pair of axes. The exponential rises from just above the horizontal axis on the left to steeply upward on the right, never touching that axis. The logarithm climbs away from the vertical axis at the bottom and flattens to the right, never touching that axis. A dashed diagonal line y equals x runs between them. A dot at (0, 1) on the exponential and a dot at (1, 0) on the logarithm are joined by a short dashed segment that meets the diagonal at a right angle and is cut in half by it.(0, 1)(1, 0)y = bxy = logb xy = x
    Text description

    The exponential and the logarithm are mirror images across the line y = x, so their points, intercepts, and asymptotes all swap coordinates.

    Use when b>0b > 0 with b1b \ne 1; increasing when b>1b > 1, decreasing when 0<b<10 < b < 1. Domain and range are the exponential's, swapped.

Problem types, step by step

Add or subtract two polynomials

  1. For a subtraction, distribute the minus across every term of the second polynomial first.
  2. Drop the parentheses and group the terms by power.
  3. Add the coefficients in each group, leave the power unchanged, and write descending order.

e.g. (6x2x)(2x2+4x3)=6x2x2x24x+3=4x25x+3(6x^2 - x) - (2x^2 + 4x - 3) = 6x^2 - x - 2x^2 - 4x + 3 = 4x^2 - 5x + 3.

Expand a product of two polynomials

  1. Count the products you owe: mm terms times nn terms is mnmn of them.
  2. Multiply every term of the first by every term of the second, carrying signs, coefficients multiplying and exponents adding.
  3. Use a grid past four products, so none is dropped.
  4. Combine like terms and write standard form.

e.g. (x+2)(x23x+1)=x3x25x+2(x + 2)(x^2 - 3x + 1) = x^3 - x^2 - 5x + 2, from six products.

Decide whether a rule or a table is exponential

  1. In a rule, locate the variable: in the exponent is exponential, in the base is a power function.
  2. In a table with equally spaced inputs, test the differences first; a constant difference is linear.
  3. Then test the ratios: a constant ratio is exponential, and when the inputs step by 11 that ratio is the base bb.
  4. Read aa as the output at x=0x = 0 and write f(x)=abxf(x) = a \cdot b^{x}.

e.g. Outputs 2,6,18,542, 6, 18, 54 at x=0,1,2,3x = 0, 1, 2, 3: ratio 33, so f(x)=23xf(x) = 2 \cdot 3^{x}.

Describe and evaluate f(x)=abxf(x) = a\,b^{x}

  1. With a>0a > 0, read the base for growth (b>1b > 1) or decay (0<b<10 < b < 1), and aa as the value at x=0x = 0.
  2. To evaluate, substitute: a zero exponent gives 11, a negative exponent a reciprocal, a fractional exponent a root.
  3. To describe, add the asymptote y=0y = 0 and, for a>0a > 0, the range y>0y > 0.

e.g. f(x)=24(12)xf(x) = 24\left(\tfrac{1}{2}\right)^{x} decays from 2424, and f(3)=2418=3f(3) = 24 \cdot \tfrac{1}{8} = 3.

Find a compound-interest balance

  1. Convert the percent to a decimal and read off PP, rr, tt, and nn.
  2. Substitute, dividing the rate by nn and multiplying the exponent by nn.
  3. Evaluate the growth factor in full, multiply by PP, and round only the final balance to the nearest cent.

e.g. 20002000 dollars at 4%4\% compounded annually for 22 years: 2000(1.04)2=2163.202000(1.04)^{2} = 2163.20 dollars.

Evaluate a logarithm by hand

  1. Ask what the definition asks: the base to what power gives the input?
  2. Rewrite the input as a power of the base, using a negative exponent for a reciprocal.
  3. Report that exponent. If the input is 00 or negative, stop: the logarithm is undefined.

e.g. log2 ⁣(116)\log_{2}\!\left(\tfrac{1}{16}\right): since 24=1162^{-4} = \tfrac{1}{16}, the answer is 4-4.

Expand or condense with the laws of logarithms

  1. Factor the input into pieces whose logarithms you know or were given.
  2. Split products into sums and quotients into differences, one base throughout.
  3. Bring every exponent on an input down to a coefficient in front.
  4. To condense, run the laws backward: coefficients become exponents, sums one product, differences one quotient.

e.g. Given logb(2)=0.30\log_{b}(2) = 0.30 and logb(3)=0.48\log_{b}(3) = 0.48: logb(12)=2(0.30)+0.48=1.08\log_{b}(12) = 2(0.30) + 0.48 = 1.08.

Solve for an unknown exponent, including a doubling time

  1. Isolate the power so the equation reads bx=yb^{x} = y with y>0y > 0; doubling gives (1+r)t=2(1 + r)^{t} = 2.
  2. Take the same logarithm of both sides and use the power law: xlog(b)=log(y)x\log(b) = \log(y).
  3. Divide by log(b)\log(b) and evaluate.

e.g. Doubling at 5%5\% compounded annually: t=log2log1.05=0.30100.021214.2t = \frac{\log 2}{\log 1.05} = \frac{0.3010}{0.0212} \approx 14.2 years.

Exam traps

  • Trap Splitting the logarithm of a SUM: log2(4+4)=log2(4)+log2(4)=4\log_{2}(4 + 4) = \log_{2}(4) + \log_{2}(4) = 4.

    Fix No law breaks up the log of a sum. Add inside first: log2(4+4)=log2(8)=3\log_{2}(4 + 4) = \log_{2}(8) = 3. The product law needs a PRODUCT.

  • Trap Flipping only the first term of a subtraction: (5x23x+4)(2x2+6x7)=3x2+3x3(5x^2 - 3x + 4) - (2x^2 + 6x - 7) = 3x^2 + 3x - 3.

    Fix All three signs flip: 5x23x+42x26x+7=3x29x+115x^2 - 3x + 4 - 2x^2 - 6x + 7 = 3x^2 - 9x + 11.

  • Trap Squaring a sum term by term: (x+4)2=x2+16(x + 4)^2 = x^2 + 16.

    Fix (x+4)2(x + 4)^2 is (x+4)(x+4)(x + 4)(x + 4), four products with two cross terms: x2+8x+16x^2 + 8x + 16.

  • Trap Doing the wrong thing to exponents: 3x2+5x2=8x43x^2 + 5x^2 = 8x^4 when adding, x2x3=x6x^2 \cdot x^3 = x^6 when multiplying.

    Fix Adding like terms adds the COEFFICIENTS and leaves the power alone, giving 8x28x^2. Multiplying powers ADDS the exponents, giving x5x^5.

  • Trap Reading b0b^{0} as 00, or a negative exponent as a negative output, so 232^{-3} looks like 8-8.

    Fix b0=1b^{0} = 1 for every allowed base, and a negative exponent gives a reciprocal: 23=182^{-3} = \tfrac{1}{8}, positive like every exponential output.

  • Trap Treating bxb^{x} and xbx^{b} as the same species.

    Fix 2x2^{x} is exponential, x2x^{2} a power function: at x=3x = 3 they give 88 and 99, and past x=4x = 4 the exponential outruns it for good.

  • Trap Putting the percent itself into an interest formula: A=1000(1+5)tA = 1000(1 + 5)^{t} for 5%5\%.

    Fix The rate is the decimal, so A=1000(1.05)tA = 1000(1.05)^{t}. Writing 55 multiplies the balance by 66 every year.

  • Trap Solving bx=yb^{x} = y as x=log(y)log(b)x = \log(y) - \log(b), or as x=log ⁣(yb)x = \log\!\left(\tfrac{y}{b}\right).

    Fix Both are the quotient law, a different situation. Dividing xlog(b)=log(y)x\log(b) = \log(y) by log(b)\log(b) leaves x=log(y)log(b)x = \dfrac{\log(y)}{\log(b)}, a QUOTIENT.

Chapter test Questions from across the chapter