Unit+5

Given the function g(x) = x^2 - 7 when x<=0 1.) Find the inverse algebraically and explain your step-by-step process verbally. You would find the inverse by replacing, in the original equation, where x is with y, and where y is with x. So, it would look like x = y^2 – 7. Then, solve for x. Add 7 to each side first, so it would be x + 7 = y^2, then take the square root of each side so it would be the plus or minus the root of (x + 7) = y. But since x is less than or equal to 0, it would just be the negative the root of (x+7) = y.
 * Part 1 **

2.) The graph of g(x) is labeled below. Explain how you can find at least 4 coordinate points that represent the inverse function. Write the coordinate points you found. I can find at least 4 coordinate points that represent the inverse of the function by finding 4 points on the original function and switching the positions of the x and y in the coordinates, so the x would be y, and the y would be x. Four coordinates would be (-7, 0), (-3, -2), (2, -3) and (9, -4).

3.) Explain why g(x) was given the domain x<=0 rather than sketching the entire function when finding the inverse graphically.  G(x) was given the domain x is less than or equal to 0 so it would also sketch out the graph where the x is less than or equal to 0. And when finding the inverse, it would have its range restricted to less than or equal to 0.

4.) Given the domain of a function h(x) is [-3, infinity) and the range is [0, infinity), find the domain and range of the inverse function. Explain how you arrived at your answer. The domain of the inverse function would be [0, infinity) and and the range would be [-3, infinity). I arrived at my answer by putting what originally was the domain of h(x) as the range of the inverse function, and what originally was the range of h(x) as the domain of the inverse function.

 **__ Part 2 __** 1) Explain the step by step process of graphing the function g(x) = 1 - 2log3(x+4) (this is a log base 3, the 3 should not be distributed through the parenthesis) without using the graphing calculator. Find three point of the parent function and the three corresponding points of g(x). List the asymptote(s), domain, and range. To graph the function g(x), use its inverse, 3^x, and find 3 coordinate points, so (0, 1), (1, 3), and (2, 9). Then switch the positions of x and y so that it would be usable for log3. The coordinates would be (1,0), (3,1), and (9,2). From there, know that you will be moving 4 left, reflecting over the y axis, stretching by 2, and moving 1 up. So, do those steps to all the coordinate pairs, and the resulting points should be (-3, 1), (-1, -1), and (5, -3).

2.) Explain how you can find the domain and range of any logarithmic function without looking at the graph or using a graphing calculator. The domain of any log function could be found by setting whatever is next to the log base, so in the case above, whatever is in the parenthesis, greater than 0 and solve for x. The range will always be all real numbers.

Evaluate the following expression. Since I cannot control whether you use a calculator at home you must write your steps out verbally so I know you understand the process. <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;"> A) First, using the rules of expanding logs, expand the expression, so it would be ln 1 – (2/5)ln e. After, know that e^0 is equal to 1, so ln 1 would be 0 and for the second part, ln e would equal 1, multiplied by (2/5) would be (2/5). So, 0 – (2/5) would be -2/5. <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">B) In this expression, know that the fourth root of 27 is the same as 27^1/4. Using the rules of expanding logs, know that the expression can be transformed into (1/4)log327. Know that 3^3 is equal to 27, so log327 would be equal to 3 and multiplied by (1/4) would be (¾). <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">C) Expand the expression using the rules of expanding logs so it would be 3log216. Know that 2^4 is equal to 16 so log216 = 4 and multiplied by 3 equals 12. <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">D) Expand this expression further, making it 2ln e – 3ln e. ln e = 1, so knowing this fact, it would make the expression 2(1) – 3(1) and simplifying that further would be -1. <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">E) Using the rules of expanding logs, expand the expression further making it log41 – log416. Know that 4^0 = 1, and that 4^2 is 16, so log41 = 0 and log416 = 2. The expression would then be 0 – 2, simplifying to -2. <span style="background-color: white; color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">F) Simplify the fraction to 1/125, and use the rules of expanding logs to expand the expression, making it log51 – log5125. Then, knowing that 5^0 = 1 and 5^3 = 125, log51 would be 0 and log5125 would be 3. The expression would then be 0 – 3, simplified further to -3.
 * __<span style="background-color: white; font-family: 'Century Schoolbook',serif; font-size: 14pt;">Unit 5 Lesson 5 __**

<span style="font-family: 'Century Schoolbook',serif; font-size: 14pt;">In Unit 5 Lesson 6 We learned about rewriting logarithmic expressions by expanding to have multiple logarithms and condensing to have a single logarithm.I want you to prove algebraically, why the following statements are true using properties of logarithms. <span style="font-family: 'Century Schoolbook',serif; font-size: 14pt;">A) <span style="background-color: white; font-family: 'Century Schoolbook',serif; font-size: 14pt;">This is true. Log5(1/250) can be expanded to -3 – log52. To expand log5(1/250), know that a fraction means that it will be subtracted so log51 – log5250 would be the result. Then, know that log51 = 0 because 5^0 =1, so it would be 0 – log5250. But this can further be simplified because 250 can be obtained through (125)(2), so using the rules of expanding logs and distributing the negative sign, it would be –log5125 – log52. Know that –log5125 = 3 because 5^3 = 125, so it would be -3 – log52, making that statement above true.  <span style="font-family: 'Century Schoolbook',serif; font-size: 14pt;">B) <span style="background-color: white; font-family: 'Century Schoolbook',serif; font-size: 14pt;">This is true. –(3ln2 + ln3) can be condensed to –ln24. To condense the expression, know that if there is a number before the ln, it means that is the exponent of the number after it. So it would be –(ln2^3 + ln3) and can be simplified to –(ln8 + ln3). After, know that addition means to multiply so it would be –(ln(8)(3)) which can be simplified to –ln24, proving the statement correct.
 * __<span style="background-color: white; font-family: 'Century Schoolbook',serif; font-size: 14pt;">Unit 5 Lesson 6 __**

<span style="font-family: 'Century Schoolbook',serif; font-size: 14pt;">After break we will be learning how to solve various logarithmic equations and how to do applications algebraically; however if you understand the concept of exponential applications your should be able to solve and analyze a logarithmic application. <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">A) The initial amount of drugs that was administered should be when t = 0, meaning when it was 0 hrs. So, plugging in 0 to f(x), it would be 90 – 52ln(1+0) = 90 – 52(0) = 90. The amount of initial drugs was 90 mg. <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">B) To solve this problem, know that you will be setting 66 equal to 90 – 52ln(1+t). Then, solve for t, so: <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">After .5865 hrs, the amount of drugs will be 66mg. <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif;">
 * __<span style="background-color: white; font-family: 'Century Schoolbook',serif; font-size: 14pt;">Unit 5 Lesson 7 __**
 * <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">66 = 90 – 52ln(1+t)
 * <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">-24 = -52ln(1+t)
 * <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">24/52 = ln(1+t)
 * <span style="color: #ff00ff; font-family: 'Century Schoolbook',serif; font-size: 14pt;">Put into calculator, set 24/52 equal ln(1+t), then using the 2nd Trace feature, find the intersection, which the x would be the amount of hrs.