Exponential functions have inverses. The inverse of an exponenetial function is called a logarithmic function. To graph a logarithmic function, reflect the exponential graph through the line y=x on a standard graph. The x and y values switch [(1,3) becomes (3,1)]. Since an exponential function lies in quadrants I and II, the logarithmic functions will lie in quadrants I and IV.
The form of a logarithmic function is y = log b x, where b is the base, y is the exponent, and x is the result of raising b to the y power. In other words, y = log x b is the inverse of by = x. Therefore, when we rewrite y = log 2 16 in exponential form, we write: 2y = 16. When we rewrite 34 = x, we write 4 = log 3 x. To solve for 3 = log 5 x, we rewrite it as an exponent: 53 = x, and solve. x = 125.To solve x = log 2 64, we do the same: 2x = 64, and x = 6. To solve 2 = log x 9, we do the same: x2 = 9, and x=3.

 Just like exponents, logarithms have rules.

1) product rule - we add the logarithms when we see multiplication of the same base.

2) quotient rule - we subtract the logarithms when we see division of the same base.

3) power rule - we multiply the logarithm by the power that is shown when we see this.

Let's look at examples. If log _{b ^{3 = a and log} b} 4 = c, find log _{b ^{12. 12 is 4 * 3, so log} b} 12 = log _{b ^{3 + log} b} 4. Substituting values, log _{b ^{12 = a + c. Find log} b} 36. 36 is 12 * 3. We saw in the previous example that 12 is log _{b ^{3 + log} b} 4; therefore, log _{b ^{36 is a + c + a, or 2a + c.}}

If log b 5 = 1.42, find log b 25. 25 is 52. Therefore, log b 25 is 2 log b 5. 2(1.42) = 2.84

Rewrite log b w = log b x + log b y - .5 log b z as a single expression. Using the three rules above, w = xy/radical z.

Solve: log 2 x - log 2 4x = 6. using the rules above, log 2 (x * 4x) = 26. log 2 4x2 = 64, or 4x2 = 64. Solve:

4x^{2 _{= 64 Divide by 4: x}2} = 16, x = 4, -4 All logarithms are positive numbers so x = 4.

If a logarithm does not have a number written as the base, the base is automatically 10. For base 10 logarithms we can use our calculators. Just simply punch in log followed by the number and then enter. The answer is the logarithm. If we have the logarithm, punch in 2nd log followed by the number and then enter.

To solve logarithmic equations, rewrite the equation as an exponential equation and take the log of both sides. Then solve.

5^{x _{= 13 => x log 5 = log 13 => x = (log 13)/(log 5) => x = 1.59}}

x^{6 _{= 34 => 6 log x = log 34 => log x = (log 34)/6 => log x = .2552 => x = 1.79}}

_{^{log (x - 1) + log (2x - 3) = 1}}

^{_{(x - 1)(2x - 3) = 10}1} => 2x^{2_{- 2x - 3x + 3 = 10}}

_{2x² - 5x + - 7 = 0}

_{^{(2x - 7)(x + 1) = 0 => 2x - 7 = 0 ; x + 1 = 0 => x = 7/2 ; x = -1}}

_{^{Since logarithms can not be negative numbers, the answer is x = 7/2.}}

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_{^{To do regressions, we need a graphing calculator and punch the data into our calculator. The chip inside the calculator does all the math for us. All we need to do is select the regression we want and hit enter. The calculator does the rest.}}

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We could use trigonometry in various real life situations. All we need to do is use either the law of cosines or the law of sines, and we are ready to go. Both of these laws are used with triangles. If we have two sides of a triangle and the angle between them and we are looking for the third side, we use the Law of Cosines:

c^{2 _{= a}2} + b^{2 _{- 2ab(cos C)}}

If side a = 5 meters and side b = 3 meters and the angle between is 17 degrees, we would plug them into the formula:

c^{2 _{= 25 + 9 - 30 (-.2751633381), or c}2} = 34 - 8.254900142, or c^{2 _{= 25.74509986, or c = 5.07 meters.}}

_{^{If we have the measures of all three sides and we are asked to find an angle, we use the formula:}}

^{_{cos C = ( a}2} + b^{2 _{- c}2_{)/2ab   [WE ALWAYS SUBTRACT THE SIDE OPPOSITE THE ANGLE WE ARE LOOKING FOR]}}

_{^{If we have a triangle with sides of 4 meters, 7 meters, and 9 meters, and are asked to find the smallest angle, we would:}}

_{^{cos C = (49 + 81 - 16)/ 126, or 114/126, or .9047619048. The angle would be 25.2 degrees.}}

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We use the Law of Sines to find the area of a triangle that is not a right triangle or whose height is not inside the triangle.

The Law of Sines is: (sin A/a) = (sin B/b) = (sin C/c).

For example, if side a = 5 yards and side b = 10 yards and angle A is 14 degrees, find angle B.

(sin (14)/5) = (sin B/10), or 2.419218956 = 5sin B, or .4838437912 = sin B, or B =28.9 degrees.

Remember the following rules for the Law of Sines:

1) if the given angle is obtuse and side a is less than or equal to side c, 0 triangles can be formed.

2) if the given angle is obtuse and side a is greater than side c, 1 triangle can be formed.

3) if the given angle is acute and side c is greater than side a is greater than c*sin A, 2 triangles can be formed.

4) if the given angle is acute and a = c*sin A, 1 triangle can be formed.

5) if the given angle is acute and a is greater than or equal to side c is greater than or equal to c*sinA, 1 triangle can be formed.   

6) if the given angle is acute and a is less than c*sin A, 0 triangles can be formed.

One more note here: we can use the Law of sines or Cosines to solve force and vector problems or triangle area problems.

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Sometimes we are asked to rewrite angles as combinations of other angles. In these cases we need to express them as best we could in terms of special angles (30,45,60,90) since we know these function values. There are formulae we use in these cases. They are:

SUM OF ANGLES - sin (A + B) = (sin A * cos B) + (cos A* sin B)

                                   cos (A + B) = (cos A* cos B) - (sin A * sin B)

                                   tan (A + B) = (tan A + tan B)/(1 - tan A * tan B)

DIFFERENCE OF ANGLES - sin (A - B) = (sin A * cos B) - (cos A * sin B)            

                                                 cos (A - B) = (cos A * cos B) + (sin A * sin B)

                                                 tan (A - B) = (tan A - tan B)/(1 + tan A * tan B)

In addition we may be required to double certain angles or halve certain angles. In these cases we use the following formulae:

DOUBLE ANGLES - sin 2A = 2 sin A *cos A

                                   cos 2A = cos ^{2 _{A - sin} 2} A, or 2 cos^{2 _{A - 1, or 1 - 2 sin} 2} A (depending on what information we are given)

HALF ANGLES - sin .5A = +/- the square root of [(1-cos A)/2]

                              cos .5 A = +/- the square root of [(1+cos A)/2]

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We may need to solve equations with trigonometric functions. These could be first degree, second degree, or a combination of degrees. To solve first degree equations, it is a matter of thinking of the problems as if the function is any other variable. Once we solve for the function, we can find the angle. Most of these equations will have two answers since all functions are positive or negative in two quadrants. For example:

4 sin x + 3 = 5

4 sin x = 2

sin x = 2/4, or 1/2

x = 30 degrees or 150 degrees

We need to think of second degree equations as quadratic equations that need to be factored first and then solved. For example:

sin^{2 _{x - sin x - 2 = 0}}

_{^{(sin x - 2) (sin x + 1) = 0}}

_{^{sin x = 2, -1 BUT SINCE SINE IS NEVER LARGER THAN 1, THE ANSWER IS -1}}

_{^{x = 270 degrees}}

_{^{Sometimes the quadratic equation is needed to solve these problems.}}

_{^{If there is more than one function in the equation, we need to use the identities to convert one function into another. for example:}}

^_cos2 x + 2 sin x = 1                 Using the Pythagorean identities, cos^{2 _{x = 1 - sin}2} x

1 - sin^{2 _{x + 2 sin x = 1}}

^{_-sin 2} x + 2 sin x = 0

sin x (-sin x + 2) = 0

sin x = 0, 2 BUT WE MUST REJECT 2 - SEE ABOVE

sin x = 0 and x = 0 degrees, 180 degrees, 360 degrees

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To perform proofs, we need to be familiar with certain definitions. All these definitions are in your workbook on pages 1-3. We also need to be familiar with certain relationships. For example, two triangles are congruent if any of the following hold true:

1) SSS - all three sides of one triangle are congruent to all three sides of another triangle

2) SAS- two sides and the included angle of one triangle are congruent to the same on another triangle

3) ASA - two angles and a side of one triangle are congruent to the same on another triangle

These are used in direct geometric proofs.

Once we prove two figures congruent, any remaining side or angle is automatically congruent because of CPCTC (Congruent Parts of Congruent Triangles are Congruent). If two triangles are to be proven similar, we need only prove two congruent angles and set any sides we are asked to in proportion to each other. When working with parallelograms, rectangles, squares, or rhombuses, it is easiest to first split them into triangles and work from there.

Sometimes we are asked to prove that a quadrilateral or three sdied figure is a specific shape. For this we use analytic proofs. Graph the figure, and use any of the following formulae in your work:

slope - (Y₂-Y₁)/(X₂-X₁)          midpoint - (Y₂+Y₁)/2;(X₂+X₁)/2         distance - radical (Y₂-Y₁) + (X₂-X₁)

For circle proofs, we need more information. This information is in our workbook on page 43.                                                                                          

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Imaginary numbers are numbers that do not exist, such as square roots of negative numbers. We represent imaginary numbers with the letter "i" and perform the indicated operation as usual. For example, the square root of -49 would be 7 because the square root of 49 is 7 but we attach an "i" to represent that it is an imaginary number. Therefore, the square root of -49 is 7i. The square root of -18 is different, though, because 18 is not a perfect square. We use prime factorization which is breaking down 18 into prime numbers. 18 can be rewritten as 3*3*2. Since there are a pair of 3's, we remove them and write them one time in front of "i" followed by the rest of the numbers under the radical sign. The square root of    -18 is 3i(radical 2).

There are only 4 powers of i that exist. All other powers are equivalent to these. i^0 or any multiple of 4 =1, i^1 or any other power that has a remainder of 1 when divided by 4 = i, i^2 or any other power that has a remainder of 2 when divided by 4 = -1, and i^3 or any other power that has a remainder of 3 when divided by 4 = -i.

To add or subtract imaginary numbers, we must break down the imaginary unit into one of the 4 above powers and solve from there, combining any like terms. For example, 4(radical -25) + 2(radical -36) = (4*5i) + (2*6i) = 20i + 12i = 32i. Another example: 4(radical -12) - 5(radical -48). 12 is 2*2*3 and 48 is 4*4*3 so we will pull out the 2 and 4, leaving behind radical 3 in both cases. Therefore, we have: (4i*2*radical 3) - (5i*4*radical 3) = 8i(radical 3) - 20i(radical 3) =     -12i(radical 3).

When multiplying or dividing, we perform the same steps as above, but we don't have to worry about like terms because rules of exponents come into play here. For example, (2*radical -50)(-2*radical -27). 50 can be broken down into 5*5*2 and 27 can be broken down into 3*3*3. Therefore, we remove the 5's and a pair of 3's, leaving the rest under the radical signs. We have: (2i*5*radical 2)(-2i*3*radical 3) = (10i*radical 2)(-6i*radical 3) = (-60i^{2_{*radical 6)}} = (-60)(-1)(radical 6) = 60(radical 6). Another example: (6*radical -300)/(2*radical -100). 300 can be broken down into 10*10*3 and 100 can be broken down into 10*10. In both cases, we remove 10's and leave what was left. We have: (6i*10*radical 3)/(2i*10) = (60i*radical 3)/(20i), which breaks down into 3(radical 3).

When we solve radical equations, we square both sides of the equation after combining all like terms. We already solved these equations when we worked with exponential equations.

All imaginary numbers have two pieces - the real part (the number without the i) and the imaginary part (the number with the i).

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Keep in mind the following rules about adding or subtracting complex numbers: Combine like terms and simplify. In the case where you are asked to find values of a and b that make an equation true, combine like terms. For example, in the case of :

(a + bi) + (7 - 3i) = 12 + 8i, combine 7 and 12 as well as -3i and 8i. After doing this, the real value is a and the imaginary value is b. Therefore, a = 5 and b = 11i.

When multiplying complex numbers, perform FOIL and combine like terms. Remember to change powers of i according to what we have said above and combine all like terms. For example:

(3 + 4i) * (-2 - 3i)

-6 - 8i - 9i - 12i^{2 _{[i ^2 = -1]}}

_{^{-6 - 12(-1) - 17i = - 6 + 12 - 17i = 6 - 17i}}

_{^{When dividing complex numbers, multiply the numerator and denominator by the conjugate of the denominator. The conjugate is the same as the denominator, except we change the sign before the imaginary part. For example:}}

_{^{(4 - 3i) / (2 + 2i) The conjugate is 2 - 2i so we multiply both pieces of this fraction by that number and combine like terms.}}

^{_{(4 - 3i)(2 - 2i) / (2 + 2i)(2 - 2i) = (8 - 6i - 8i + 6i}2_{) / (4 - 4i}2}) = (8 - 6 - 14i) / (4 + 4) = (2 - 14i)/8 = (2/8) - (14i/8) = (1/4) - (7/4i)

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To find the absolute value of complex numbers, we use the formula:

radical (a^{2 _{+ b}2}) WE IGNORE THE I WITH THESE PROBLEMS!!!

For example, the absolute value of 8 - 3i = radical (8^{2 _{+ 3}2}) = radical (64 + 9) = radical 73

If we can, we simplify the radical using the rules of radicals.

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Probability is the determination of how often a certain situation is likely to occur if we repeated the same situation repeatedly. We can solve probability problems using binomial exapansion. As we will see, binomial expansion is the premise for combinations. Binomial expansion is raising a binomial to successive powers. For example, raising (a + b) to the following powers:

(a+b)^0 = 1

(a+b)^1= 1a + 1b

(a+b)^2= 1a^2 +2ab + 1b^2

(a+b)^3= 1a^3 + 3a^2b^1 + 3a^1b^2 + 1b^3

(a+b)^4= 1a^4 + 4a^3b^1 + 6a^2b^2 + 4a^1b^3 + 1b^4

(a+b)^5= 1a^5 + 5a^4b^1 + 10a^3b^2 + 10a^2b^3 + 5a^1b^4 + 1b^5

Notice a pattern? The coefficients are important. Look at them again:

                          1

                    1       1

               1      2        1

          1      3         3      1

     1      4       6         4       1

1      5       10     10        5      1

Each line has a 1 on the outside. The second number in each line is the power that the binomial was raised to, and each number in a line under a previous line is the sum of the two numbers directly above it. For example, in line 4, we have 1's on the outside, 3 is the power for the binomial because 3 is the second number, and 3 is the result of adding 1+2 above it. We can rewrite these numbers using combinations, which is the basis of this triangle (called Pascal's Triangle):

                                                                           0C0

                                                                 1C0          1C1

                                                         2C0          2C1          2C2

                                               3C0          3C1           3C2          3C3

                                     4C0             4C1         4C2          4C3           4C4

                         5C0             5C1             5C2         5C3          5C4            5C5

We need to know this because we may be asked to find the result of a specific term in the expansion. To do this, we use the formula:

r^{th _{term of (a+b)}n} = _n^Cr-1 (a) ^{n-(r-1)_(b)r-1}

^{_{For example, find fourth term of the expansion (3x+2)}5}  To proceed, make a list of all the variables we need.

a = 3x, b = 2, n = 5, r =4, r -1 = 3 Now we can substitute these values into the formula:

5C3 (3x) ^{5 - 3 ₍₂₎ 3} = 10(9x^{2_{)(8) = 720x}2}

For example, find the middle term of the exapnsion (2m-5)^{6 _{To proceed, do as above:}}

_{^{a = 2m, b = -5, n = 6, r = 4, r-1 = 3 [the middle term is 4 because line 6 has 7 terms in the expansion]}}

^{_{6C3 (2m)}6-3_(-5)3_{= 20 (8m)}3 _{(-125) = 20,000m}3}

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_{^{To figure out the probability of exactly a certain number of successes in a certain situation (say, how many heads will appear after 4 coin tosses), we make a list of all the possibilities and write the response as a fraction. The combinations of tossing heads or tails after four tosses are listed below:}}

_{^{H,H,H,H           H,H,T,T        T,T,T,T       T,T,H,H}}

_{^{H,H,H,T          H,T,T,H         T,T,T,H       T,H,H,T}}

_{^{H,H,T,H          H,T,H,T        T,T,H,T       T,H,T,H}}

_^H,T,H,H       H,T,T,T    T,H,T,T   T,H,H,H

The probability of getting 0 heads = 1/16

The probability of getting exactly 1 head = 4/16

The probability of getting exactly 2 heads = 6/16

The probability of getting exactly 3 heads = 4/16

The probability of getting 4 heads = 1/16

Redundant to do this, isn't it? We can perform combinations to solve these types of problems. We need the probability of coin tosses for tails and heads (T + H). Since we need 4 coin tosses, we raise this binomial to the 4th power: (T + H)^4. The expansion for exactly 3 heads is:

4C3 (T) (H)^3 = 4(1/2)(1/2)^3 = 4(1/2)(1/8) = 4/16

The formula we use is: nCr (failure)^{n-r _(success)r}

^{_{We use the same formula for at most or at least situations. If we want at most 2 red spins on a three colored spinner, we use the formula:}}

3C2 (2/3)¹(1/3)^{2 _{+ 3C1 (2/3)}2 _(1/3)1 _{+ 3C0 (2/3)}3 _(1/3)0  _{= 3(2/3)(1/9) + 3(4/9)(1/3) + 1(8/27)(1) = 26/27}}

_{^{If we wanted at least 2 out of 3 heads, we would do:}}

^{_3C2(1/2)1_(1/2)2 _{+3C3 (1/2)}0_(1/2)3 _{= 3(1/2)(1/4) + 1(1)(1/8) = 1/2}}

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Statistics often requires a sum of the data that was collected over tome. In these cases, we use sigma notation. Sigma is a Greek letter. To solve problems using sigma, we add all the data. It's that simple. For example:

Evaluate: (3i - 2) from i = 4 through 7. We substitute 4,5,6,7 into this problem and add all results.

[3(4) - 2] + [3(5) - 2] + [3(6) - 2] + [3(7) - 2] = 10 + 13 + 16 + 19 = 58

We can perform these problems on our calculators by opening "2nd" "Stat" "Math" "5" "enter" "2nd" "Stat" "OPS" "5" "enter" followed by the equation, substituting x for the variable. [IN THIS CASE, WE TYPE 3X-2, X, 4, 7,1))]

All statistics use mean (average), median (middle number when placed in numerical order), and mode (most frequent number). When taking a collection of data, it is important to avoid bias (favoring one group over another to reach a desired result). Rather, we should try to collect a random sample so that the result is not predetermined.

All statistics also use a range, which is the difference between the highest and lowest values in the set. We could also use the mean standard deviation, which is the mean of the absolute values of how much each piece deviates (differs) from the mean. The formula is the summation of the absolute value of individual points minus the mean divided by the number of values in the set. For example, recent test scores in a junior science class for a particular student are: 68, 87, 98, 86, 86. We have five values in the set. The mean is 85 (add and divide by 5). Each value minus the mean produces differences of -17,2,13,1,1. The absolute values for these are 17,2,13,1,1. Add these and you get 34. Divide this by 5 and you get 6.8 This means that these scores deviate from the mean by 6.8 one way or the other.

We could do the same type of information as the previous example and use the variance this time. The variance requires that we first use the square of the differences between each point and the mean without worrying about absolute value. Test scores of 68, 87, 98, 86, 86 produce a mean of 85. The differences are -17, 2, 13, 1, 1. Square these and you get 289, 4, 169, 1, 1. Add these and you get 464 and divide by 5 for a variance of 92.8 The problem here is that we are not working with the scores themselves but their squares. To fix that, we take the square root of the variance and arrive at the standard deviation. The square root of 92.8 is 9.63

We can use our calculkators for all these. Press "Stat" "1" "Clear" "enter" and enter all your scores in list 1 and enter the frequency of each score in list 2. If a score appears one time, the frequency is 1 but, if it appears five times, the frequency is 5. Press "Stat" ">" "1" "enter" "2nd" "1" "," "2nd" "2" "enter"

For large samples of data, we could create histograms for the data. if we do, we would notice a pattern where the histogram would rise and fall at even intervals. The mean would be the middle bar in the histogram. If we draw a line across all bars, we would see a bell shape. This bell shape is a bell curve. This is a normal distribution of data. In a normal distribution, 68.2% of all data occur within 1 standard deviation of the mean and 95.4% occur within 2 standard deviations of the mean.  About 99.8% of all data occur within 3 standard deviations of the mean. We use this knowledge to divide all data into percentiles. A percentile is a point below which everyone else ranks compared to you. For example, the 75th percentile means that 75% of people rank lower than you. Normal distributions see 50% of all data below the mean and 50% of all data above the mean.

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When we graph, it is important to know if the line we are graphing is a relation or a function. A relation is any ordered set of points. A function is a line whereby each x-value has only one y-value and only one value. We could test to see if a line is a function by drawing a vertical line. If the vertical line touches the graphed line once, the graphed line is a function. We could write functions using notation: f(x) = 4x-5. This is read as, "F of x is 4x-5". 4x-5 is the line that we are to graph. If we want to find the point on the line where x=3, we simply substitute 3 into the equation and solve. f(3) = 4(3)-5 = 7. All possible x-values that satisfy a function is called the domain while all y-values that satisfy a function is called the range. We could compose functions, that is perform mathematical operations on functions if we know the two functional equations. This is basically performing operations with binomials, which we had done earlier in the year. All functions have inverses. In an inverted function, we simply swap the domain and range of both. The inverse of the function whose domain is (3,4,5,6) and whose range is (-1,-2,-3,-4) is a function whose domain is (-1,-2,-3,-4) and whose range is (3,4,5,6).

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Certain figures, when graphed, can be reflected through a single point or line so that half of it is on either side. For example, a heart, a circle, MOM all have reflections - either point or line. We could shift figures on a graph, preserving its shape. This shifting is a translation. Dilations are when we shrink or expand a figure beyond its initial size. When we use transformations (images of figures) that preserve distance, we call it an isometry. All isometries, when we perform two or more, are equivalent to a roatation or a translation of some sort.

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When we graph circles, we could disect a circle into separate pieces. The equation of a circle is x² + y² = r², where x and y are the center of the circle and r is the radius. A parabola is a function that represents all points equidistant from a central location. An ellipse is a cross-section of a cone where all radii are not equal. There are two foci in an ellipse where the ellipse begins to bend. There is a minor axis and a major axis. When we graph two parabolas side by side and they come from the same cross-section of the same cone, we call them hyperbolas. It is not unusual for a hyperbola and an ellipse to exist in the same graph. All ellipses have eccentricity, which is the flatness of the ellipse. When two figures have an inverse relationship - one variable increases while the other decreases - rectangular hyperbolas are formed. We can solve systems of equations using our knowledge of these circles and their various pieces by combining like terms and solving.

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