This workshop will. Exponential Functions
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1 1 This workshop will Arapahoe Community College MAT 111 Graphing Calculator Techniques for Survey of Algebra TI-83 Graphing Calculator Workshop #12 Exponential and Logarithmic Functions 1) explore graphs of exponential functions, 2) explore graphs of logarithmic functions, 3) demonstrate how to graph logarithmic functions with base other than e or 10 using the change of base formula, 4) show methods for solving exponential and logarithmic equations. Exponential Functions Exponential functions are entered by typing in the base, then using the key <4,5>, and then entering the appropriate exponent. Parenthesis may be required around the exponent to ensure proper order of operations. The TI-83 also has two built-in exponential functions. The exponential function with base 10 is accessed with the key (2 nd <6,1>) and the exponential function with base e is accessed with the key (2 nd <7,1>). After pressing either of these keys, the exponent is entered with parentheses as appropriate. 1) Make sure your calculator is in function graphing mode (5 th line of the mode screen). 2) Select the y= menu selection. 3) Clear all functions out of the y variables. 4) Enter f 1 (x)=e x into y1= by pressing <7,1>. 6) Set your window parameters as follows 7) Graph y1.
2 2 8) Enter f 2 (x)=10 x into y2= by selecting the y= menu selection and pressing <6,1>. 9) Change the plot style of the y2 by using the arrow keys to place the cursor on the plot style icon to the left of y2=. Then press until y2 has the bold line plotting style selected. 10) Graph y1 and y2. 11) Enter f 3 (x)=2 x into y3= by selecting the y= menu selection and pressing. 12) Change the plot style of the y3 by using the arrow keys to place the cursor on the plot style icon to the left of y3=. Then repeatedly press until y3 has the dashed line plotting style selected. 13) Graph y1, y2, y3. 14) Enter f 4 (x) = 20 x and f 5 (x) = 1.1 x into y4 and y5 respectively and determine an appropriate plot style. Turn f 2 (x) and f 3 (x) off and graph f 1 (x), f 4 (x), and f 5 (x). Sketch your results.
3 3 Logarithmic Functions The TI-83 has two built-in logarithmic functions, log e (x) ln(x) and log 10 (x) log(x). These built-in logarithmic functions are accessed with the and keys. 1) Select the y= menu selection. 2) Clear all functions out of the y variables. 3) Enter f 1 (x)=ln(x) into y1= by pressing <7,1>. 4) Set your window parameters as follows 5) Graph y1. 6) Enter f 2 (x)=log(x) into y2= by selecting the y= menu and pressing <6,1>. 7) Change the plot style of the y2 by using the arrow keys to place the cursor on the plot style icon to the left of y2=. Then press until y2 has the bold line plotting style selected.
4 4 8) Graph y1 and y2. Graphing Logarithmic Functions Using a Change of Base To graph logarithmic functions with a base other than e or 10, you must use the change of base formula log log x = log x ln( x) = b ln( b) a b = a log( x) log( b) 1) Select the y= menu selection. 2) Clear all functions out of the y variables. 3) ln( x) Enter f 1 (x)= log 2 x = into y1= by pressing ln(2). 4) Set your window parameters as follows 5) Graph y1.
5 log( x) 6) Enter f 2 (x)= log 2 x = into y2= by selecting the y= menu and log(2) pressing. 7) Change the plot style of the y2 by using the arrow keys to place the cursor on the plot style icon to the left of y2=. Then press until y2 has the bold line plotting style selected. 5 8) Graph y1 and y2. 9) Trace along y1 and y2 by selecting the trace menu selection and using the arrow keys. What can you conclude about the functions in y1 and y2? 10) Turn off y1 and y2 and graph f 3 (x)=log 4 (x) and f 4 (x)=log 8 (x) using either the ln or log functions and the change of base formula. Sketch your result. Solving Exponential and Logarithmic Equations ISECT method You will now use the ISECT method explored in Workshop 8 Solving Equations Graphically to solve exponential and logarithmic equations. Remember that the x coordinates of the points of intersection of f 1 (x) and f 2 (x) are the real solutions to f 1 (x)=f 2 (x). Suppose you want to solve ln(3x+5)=2. You could graph the two functions and estimate the points of intersection using the graph math ISECT menu selection. 1) Select the y= menu selection.
6 6 2) Clear all functions out of the y variables. 3) Enter f 1 (x) = ln(3x+5) into y1. 4) Enter f 2 (x) = 2 into y2. 5) Set your window parameters as follows 6) Graph y1 and y2. 7) Select the CALC menu selection. 8) Press to select the intersect menu selection. 9) You will find the intersection point of y1 and y2. ISECT needs you to identify the two curves that have the point of intersection you are interested in and an initial guess of the point of intersection. Use the arrow keys to get the cursor somewhere on the first function (notice the y1=ln(3x+5) in the upper left corner). 13) Press. 14) Use the arrow keys to get somewhere on the second function (notice the y2=2 in the upper left hand corner).
7 7 15) Press. 16) Use the arrow keys to enter a guess of the point of intersect to start the numerical algorithm 17) Press. 18) The x coordinate of the point of intersection is a solution to ln(3x+5)=2, so the solution is approximately x= Now, suppose you want to solve 3 2x-1 =5. 1) Select the y= menu selection. 2) Clear all functions out of the y variables. 3) Enter f 1 (x) = 3 2x-1 into y1. 4) Enter f 2 (x) = 5 into y2. 5) Set your window parameters as follows 6) Graph y1 and y2.
8 8 7) Select the CALC menu selection. 8) Press to select the intersect menu selection. 9) You will find the intersection point of y1 and y2. ISECT needs you to identify the two curves that have the point of intersection you are interested in and an initial guess of the point of intersection. Use the arrow keys to get the cursor somewhere on the first function (notice the y1=3^(2x-1) in the upper left corner). 13) Press. 14) Use the arrow keys to get somewhere on the second function (notice the y2=5 in the upper left hand corner). 15) Press. 17) Use the arrow keys to enter a guess of the point of intersect to start the numerical algorithm 17) Press. 18) The x coordinate of the point of intersection is a solution to 3 2x-1 =5, so the solution is approximately x= Here are some more problems to practice what you have learned in this workshop.
9 9 log(x) + log(x+3) = 1 answer x=2 ln(x) = x 2 2x - 2 answer and 2 x 2 2 4x = 32 answer x=-5 and x=1 e x = x+2 answer and
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