Unit+V+Virtual+Responses

Given the function g(x) = x^2 - 7 when x<=0 The inverse of the function would be plus or minus the square root of x+7. This is achieved by first replacing the x-values with y-values and the y-values with x-values, which would leave the function to be x=y^2-7. Next, 7 would be added to one side and taken from the other (x+7=y^2). Then you would proceed to square root both sides to get 'y' by itself, leaving you with plus or minus the square root of x+7. But since x has to be less than or equal to, the plus option is not an option because they are positive numbers. You can find at least four coordinate points that represent the inverse function using the graph of g(x) by finding four points of the original graph and switching the 'x' and 'y'-values. For example, four points that can be seen on this graph are: (-4,9), (-3,2), (-3,-2), and (-1,-6). From finding these points, you can easily find four points that represent the inverse function - (9,-4), (2,-3), (-2,-3), and (-6,-1).  The g(x) was given the domain of x<=0 rather than sketching the entire function when finding the inverse graphically because it would show where x is less than or equal to 0 because when finding the inverse, the range would have a max of y<=0 because the x and y values switch. Domain of h(x): [0, infinity) Range of h(x): [-3,infinity) I arrived at this answer due to the fact that when creating an inverse function, the 'x' and 'y' values change.
 * Unit 5, Lesson 4 **
 * __Part 1 __**
 * 1.) Find the inverse algebraically and explain your step-by-step process verbally. **
 * 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. **
 * 3.) Explain why g(x) was given the domain x<=0 rather than sketching the entire function when finding the inverse graphically. **
 * 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. **

First when graphing the function g(x)=1-2log3(x+4) I would begin to create a table of it's inverse function, 3^x including the points: (-2,1/9), (-1, 1/3), (0,1), (1,3), and (1,9). Then after finding these points you would create an inverse for this table leaving the points to be (1/9,-2), (1/3,-1), (1,0), (3,1) and (9,2). Afterwards, you would create another table knowing that from the original functions **1-2log3(x+4)** the points would shift to the **left** 4, reflect over the **y-axis** and stretch by 2, leaving the coordinates to be: For any log function, the range will **always** be all real numbers no matter what. The domain of any log function is always what is in the parenthesis next to the log symbol with the opposite sign.
 * __<span style="background-color: white; font-family: Arial,sans-serif; font-size: 10.5pt;">Part 2 __**
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10.5pt;">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. **
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10.5pt;">2.) Explain how you can find the domain and range of any logarithmic function without looking at the graph or using a graphing calculator. **

<span style="background-color: white; font-family: Arial,sans-serif; font-size: 10.5pt;">In unit 5 lesson 5 we evaluated logarithmic and exponential expressions without using a calculator. 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; font-family: Arial,sans-serif; font-size: 10.5pt;">Unit 5 Lesson 5 __**

<span style="font-family: Arial,sans-serif; font-size: 10pt;"> <span style="font-family: Arial,sans-serif; font-size: 10pt;">a) By using the expansion rules, the expression can be expanded into ln1-2/5lne. ln1 = e^x=1. From this you can take that x=0. Then for 2/5 lne, ln and e always cancel each other out so you are left with 2/5. And then you plug in zero! <span style="font-family: Arial,sans-serif; font-size: 10pt;">b) 4th root of 27 = 27^(1/4). Using the expansion rules once again the expression can be rewritten as 1/4log3^27 <span style="font-family: Arial,sans-serif; font-size: 10pt;">c)Expansion rules again - expression can be written into 3log2^16. 2^x=16, x=4. THen 4^3=16, which is the answer. <span style="font-family: Arial,sans-serif; font-size: 10pt;">d)Expansion rules - 2lne-3lne; Ln and e cancel out meaning that the equation would be left as 2-3, which equals -1. <span style="font-family: Arial,sans-serif; font-size: 10pt;">e)Expansion rules - ln4^1-ln4^16; 4^x=1, x=0; 4^x=16 x=2; Then you taken these two answers and subtract from each other and you are left with -2. <span style="font-family: Arial,sans-serif; font-size: 10pt;">f)First, you simplify within the parenthesis to 1/125. Then you expand the expression on the outside and inside as log5^1-log5^125. Then you evaluate - 5^x=1, x=0; 5^x=125, x=3; Then you subtract once again and you are left with -3.

<span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">In Unit 5 Lesson 6 We learned about rewriting logarithmic expressions by expanding to have multiple logarithms and condensing to have a single logarithm.
 * __<span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">Unit 5 Lesson 6 __**

<span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">I want you to prove algebraically, why the following statements are true using properties of logarithms. <span style="font-family: Arial,sans-serif;">ni

<span style="font-family: Arial,sans-serif; font-size: 10pt;">a)log5(1/250) **<span style="font-family: Arial,sans-serif; font-size: 10pt;">= log5 1 - log 5 250 ** <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">log5 1 = **5^x = 1** <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">log 5 250 = 5^x = 250. **NO SOLUTION** - expand expression using the expansion rules for multiplication and addition. <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">log 5 250 **= log5^125 + log5^2** <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">log5 125 **= 5^x = 125. X=3** <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">The answer is 0 - 3 - log5 2 b) 3n2 = **ln2^3** ln8+ln3 = **ln(8*3)** =**ln(24****)**
 * <span style="font-family: Arial,sans-serif; font-size: 10pt;">log 5 2 cannot be simplified. **
 * <span style="background-color: white; font-family: Arial,sans-serif; font-size: 10pt;">= -3 - log 5 2, which proves that this statement is true. **

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.
 * Unit 5, Lesson 7**

a)The initial amount of this drug administered for pain was 90mg. b)**The amount of drug was at 66 mg at 0.5 hours.** 90-x=66: **x=24** 52ln(1+t)=24: **0<=t<=4** FINAL ANSWER: t=0.5865 hours