From b1f17b256f1031f176caebe4fbdd7689668db8b8 Mon Sep 17 00:00:00 2001 From: tdegeorge <143553140+tdegeorge@users.noreply.github.com> Date: Tue, 17 Dec 2024 00:29:28 +0000 Subject: [PATCH 01/19] FN5 addition #332 --- source/precalculus/source/02-FN/05.ptx | 132 +++++++++++++++++++++++++ 1 file changed, 132 insertions(+) diff --git a/source/precalculus/source/02-FN/05.ptx b/source/precalculus/source/02-FN/05.ptx index 5418cc38..d47cec66 100644 --- a/source/precalculus/source/02-FN/05.ptx +++ b/source/precalculus/source/02-FN/05.ptx @@ -602,7 +602,139 @@ Let functions p and q be given by the graphs below.

+ +

+ Now that we've seen how to compose functions, let's take a look at how composition of functions can be applicable in our lives. +

+ + + +

+ Suppose you have a $50-off coupon and there is a 15\%-off sale on a TV. If you are allowed to apply both the coupon and sale price of the TV, which one should you apply first? Let's investigate to determine the better deal. +

+ + + + +

+ Write a cost function, C(p), for the price of the TV, p, if you applied the $50-off coupon. +

+ + +

+ C(p)=p-50 +

+ + + + + +

+ Write a cost function, S(p), for the price of the TV, p, if you applied the 15\%-off sale. +

+ + +

+ S(p)=p-.15p +

+

+ S(p)=.85p +

+ + + + + +

+ Suppose you apply the coupon first and then the 15\%-off sale. Write a composite function to represent this situation. +

+ + +

+ S(C(p))=S(p-50) +

+

+ S(C(p))=.85(p-50) +

+

+ S(C(p))=.85p-42.5 +

+ + + + +

+ Suppose you apply the 15\%-off sale first and then the coupon. Write a composite function to represent this situation. +

+ + +

+ C(S(p))=C(.85p) +

+

+ S(C(p))=(.85p)-50 +

+ + + + + +

+ Suppose the original cost of the TV is $500. Determine the cost of the TV if you applied the coupon first. +

+ + +

+ From part (c), we know that: S(C(p))=.85p-42.5 +

+

+ S(C(500))=.85(500)-42.5 +

+

+ S(C(500))=382.50 +

+

+ When applying the coupon first, the reduced cost of the TV is $382.50. +

+ + + + + +

+ Suppose the original cost of the TV is $500. Determine the cost of the TV if you applied the 15\%-off sale first. +

+ + +

+ From part (d), we know that: S(C(p))=(.85p)-50 +

+

+ S(C(500))=(.85(500))-50 +

+

+ S(C(500))=375 +

+

+ When applying the 15\%-off sale first, the reduced cost of the TV is $375. +

+ + + + + +

+ Which is the better deal and how much would you save? +

+ + +

+ When applying the coupon first, the TV would cost $382.50. If applying the 15\%-off sale first, however, the TV would cost $375. Applying the 15\%-off sale first is the better deal, saving you $125 (versus $117.50)! +

+ + + From 33b0aee764739f2845253ab4c84a335fbde35782 Mon Sep 17 00:00:00 2001 From: tdegeorge <143553140+tdegeorge@users.noreply.github.com> Date: Tue, 17 Dec 2024 00:29:39 +0000 Subject: [PATCH 02/19] Added in an application example at the end From 04ae0f3f7b43f3041bad76b2e8742ec5a8a57784 Mon Sep 17 00:00:00 2001 From: tdegeorge <143553140+tdegeorge@users.noreply.github.com> Date: Tue, 17 Dec 2024 02:07:42 +0000 Subject: [PATCH 03/19] Fixed One-to-one property of logs in EL3, EL5, and EL6 #430 --- source/precalculus/source/05-EL/03.ptx | 18 +++-- source/precalculus/source/05-EL/05.ptx | 106 ++++++++++++++++++++++--- source/precalculus/source/05-EL/06.ptx | 26 +++--- 3 files changed, 117 insertions(+), 33 deletions(-) diff --git a/source/precalculus/source/05-EL/03.ptx b/source/precalculus/source/05-EL/03.ptx index e5f2e70f..f1c5546d 100644 --- a/source/precalculus/source/05-EL/03.ptx +++ b/source/precalculus/source/05-EL/03.ptx @@ -109,7 +109,7 @@

- Since P has an inverse function, we know there exists some other function, say L, such that y=P(t) represent the same relationship between t and y as t=L(y). In words, this means that L reverses the process of raising to the power of 10, telling us the power to which we need to raise 10 to produce a desired result. + Since P has an inverse function, we know there exists some other function, say L, such that y=P(t) represent the same relationship between t and y as t=L(y). In words, this means that L reverses the process of raising to the power of 10, telling us the power to which we need to raise 10 to produce a desired result. Fill in the table of values for L(y). @@ -188,7 +188,10 @@

- Domain (-\infty, \infty) Range (0,\infty) + Domain: (-\infty, \infty) +

+

+ Range: (0,\infty)

@@ -200,14 +203,19 @@

- Domain (0,\infty) Range (-\infty, \infty) + Domain: (0,\infty) +

+

+ Range: (-\infty, \infty)

+ - The powers of 10 function P(t) has an inverse L. This new function L is called the base 10 logarithm. But, we could have done a similar procedure with any base, which leads to the following definition. + The powers of 10 function P(t) has an inverse L. This new function L is called the base 10 logarithm. But, we could have done a similar procedure with any base, which leads to the following definition. +

@@ -223,7 +231,7 @@ \log_{b}(x)=y \iff b^{y}=x - whenever b \gt 0, b \neq 1 + whenever b \gt 0, b \neq 1 . diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 72a29089..b2bf23be 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -14,14 +14,98 @@ In this section, we will explore the properties of logarithms and learn how to manipulate them that will be helpful when we are ready to solve logarithmic equations. + + + Recall from that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x + + + + +

+ Suppose you are given two equations + \log_bM=x and \log_bN=x. +

+ + + + +

+ Rewrite each logarithmic equation into an exponential equation. +

+ + +

+ \log_bM=x \iff b^{x}=M +

+

+ \log_bN=x \iff b^{x}=N +

+ + + + + +

+ Look at the exponential equations you found in part (a). What conclusion can you make about M and N? +

+ + +

+ Because b^{x}=M and b^{x}=N, then it follows that M=N. +

+ + + + + +

+ Given that both \log_bM and \log_bN are both equal to x, what can you say about \log_bM and \log_bN? +

+ + +

+ \log_bM=\log_bN +

+ + + + + +

+ If given \log_bM=\log_bN, what can you say about M and N? (Refer back to the previous parts of this activity.) +

+ + +

+ Based on what we have seen in parts (b) and (c), we can conclude that M=N. +

+ + + + + + +

+ The one-to-one property of logarithms states that if both sides of an equation can be rewritten as a single logarithm with the same base, then the arguments can be set equal to each other. +

+ \log_bM=\log_bN if and only if M=N +

+

+ + + + +

+ Notice this definition allows us to set the arguments (M and N) equal to each other once we get each side expressed as a single logarithm with matching bases. +

+ - Recall that \log_b M=\log_b N if and only if M=N. In addition, because exponentials and logarithms are inverses, we also know that \log_b(b^{k})=k. - In addition, according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} + Recall that exponentials and logarithms are inverses. We know that \log_b(b^{k})=k and according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} \dfrac{x^a}{x^b}=x^{a-b} \left(x^a\right)^b=x^{a \cdot b} - Consider all these as you move through the activities in this section. + Consider all these as you move through the next few activities in this section. @@ -51,7 +135,7 @@

- Recall from that \log_b M=\log_b N if and only if M=N. Use this property to apply the logarithm to both sides of the rewritten equation from part a. What is that equation? + Recall from that \log_b M=\log_b N if and only if M=N. Use this property to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?

  1. \log_{10}(a+b)=\log_{10}\left(10^{x+y}\right)
  2. \log_{10}(a \cdot b)=\log_{10}\left(10^{x+y}\right)
    3. @@ -89,7 +173,7 @@

    - Recall in part a, we defined 10^x=a and 10^y=b. What would these look like in logarithmic form? + Recall in part (a), we defined 10^x=a and 10^y=b. What would these look like in logarithmic form?

    1. \log_{10}a=x
    2. \log_{x}a=10
      3. @@ -107,7 +191,7 @@

      - Using your solutions in part d, how can we rewrite the right side of the equation? + Using your solutions in part (d), how can we rewrite the right side of the equation?

      1. 10^{a+b}
      2. \log_{10}a-\log_{10}b
        3. @@ -123,7 +207,7 @@ -

        Combining parts a and d, which equation represents 10^x \cdot 10^y=10^{x+y} in terms of logarithms? +

        Combining parts (a) and (d), which equation represents 10^x \cdot 10^y=10^{x+y} in terms of logarithms?

        1. \log_{10}(a+b)=10^{a+b}
        2. \log_{10}(a \cdot b)=\log_{10}a-\log_{10}b
          3. @@ -166,7 +250,7 @@

          - Recall from that \log_b M=\log_b N if and only if M=N. Use this property to apply the logarithm to both sides of the rewritten equation from part a. What is that equation? + Use the one-to-one property of logarithms to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?

          1. \log_{10}(a-b)=\log_{10}\left(10^{x-y}\right)
          2. \log_{10}\left(\dfrac{a}{b}\right)=\log_{10}\left(10^{x-y}\right)
            3. @@ -204,7 +288,7 @@

            - Recall in part a, we defined 10^x=a and 10^y=b. What would these look like in logarithmic form? + Recall in part (a), we defined 10^x=a and 10^y=b. What would these look like in logarithmic form?

            1. \log_{10}a=x
            2. \log_{x}a=10
              3. @@ -222,7 +306,7 @@

              - Using your solutions in part d, how can we rewrite the right side of the equation? + Using your solutions in part (d), how can we rewrite the right side of the equation?

              1. 10^{a+b}
              2. \log_{10}a-\log_{10}b
                3. @@ -238,7 +322,7 @@ -

                Combining parts a and d, which equation represents \dfrac{10^x}{10^y}=10^{x-y} in terms of logarithms? +

                Combining parts (a) and (d), which equation represents \dfrac{10^x}{10^y}=10^{x-y} in terms of logarithms?

                1. \log_{10}(a-b)=10^{a+b}
                2. \log_{10}(a-b)=\log_{10}a-\log_{10}b
                  3. diff --git a/source/precalculus/source/05-EL/06.ptx b/source/precalculus/source/05-EL/06.ptx index e5d748ec..114d6df5 100644 --- a/source/precalculus/source/05-EL/06.ptx +++ b/source/precalculus/source/05-EL/06.ptx @@ -165,7 +165,7 @@

                  - Notice that the "log" has disappeared and you now have an equation with just the variable x. Which of the following is equivalent to the equation you got in part b? + Notice that the "log" has disappeared and you now have an equation with just the variable x. Which of the following is equivalent to the equation you got in part (b)?

                  1. 4x-5=2x-1
                  2. 4x-5=0
                    3. @@ -181,7 +181,7 @@

                    - Compare the answer you got in part c to the original equation given \log(4x-5)=\log(2x-1). What do you notice? + Compare the answer you got in part (c) to the original equation given \log(4x-5)=\log(2x-1). What do you notice?

                    @@ -193,7 +193,7 @@

                    - Solve the equation you got in part d to find the value of x. + Solve the equation you got in part (d) to find the value of x.

                    1. -3
                    2. \dfrac{5}{4}
                      3. @@ -213,18 +213,10 @@ Notice in , that you did not have to convert the logarithmic equation into an exponential equation. A faster method, when you have a log on both sides of the equals sign, is to "drop" the logs and set the arguments equal to one another. Be careful though - you can only have one log on each side before you can "drop" them! - - -

                      - The one-to-one property of logarithms states that if both sides of an equation can be rewritten as a single logarithm with the same base, then the arguments can be set equal to each other (and then solved algebraically). -

                      -                       -                       -

                      - Apply and other properties of logarithms (i.e., product, quotient, and power) to solve the following logarithmic equations. + Apply the one-to-one property of logarithms (see ) and other properties of logarithms (i.e., product, quotient, and power) to solve the following logarithmic equations.

                      @@ -351,7 +343,7 @@

                      - Notice that the answer you got in part b is an exact answer for x. There will be times, though, that it will be helpful to also have an approximation for x. Which of the following is a good approximation for x? + Notice that the answer you got in part (b) is an exact answer for x. There will be times, though, that it will be helpful to also have an approximation for x. Which of the following is a good approximation for x?

                      1. x \approx 5.585
                      2. x \approx 0.179
                        3. @@ -416,7 +408,7 @@

                        - Using the change-of-base formula (), rewrite your answer from part c so that x is written as a single logarithm. What is the exact value of x? + Using the change-of-base formula (), rewrite your answer from part (c) so that x is written as a single logarithm. What is the exact value of x?

                        @@ -440,12 +432,12 @@

                        - What do you notice about your answer from parts d and e? + What do you notice about your answer from parts (d) and (e)?

                        - Students should see that the two answers they got for parts d and e are the same. This might be a good time to discuss why taking the log of both sides is a valid method in solving an exponential equation. + Students should see that the two answers they got for parts (d) and (e) are the same. This might be a good time to discuss why taking the log of both sides is a valid method in solving an exponential equation.

                        @@ -572,7 +564,7 @@

                        - Compare the equation you got in part c to the original equation given. What do you notice? + Compare the equation you got in part (c) to the original equation given. What do you notice?

                        From 7a06908d75c22b1209977b5e20671ee7f09c2139 Mon Sep 17 00:00:00 2001 From: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> Date: Tue, 17 Dec 2024 15:33:13 -0500 Subject: [PATCH 04/19] Update source/precalculus/source/05-EL/05.ptx --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index b2bf23be..59a452d1 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -16,7 +16,7 @@ - Recall from that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x +

                        Recall from that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x.

                         From 7c16fa4146c83c55354f1e3fcf3ac1f388ee4ab7 Mon Sep 17 00:00:00 2001 From: tdegeorge <143553140+tdegeorge@users.noreply.github.com> Date: Tue, 17 Dec 2024 17:16:17 -0500 Subject: [PATCH 05/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 8 +++++++- 1 file changed, 7 insertions(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 59a452d1..9318852b 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -86,7 +86,13 @@

                        - The one-to-one property of logarithms states that if both sides of an equation can be rewritten as a single logarithm with the same base, then the arguments can be set equal to each other. + For any values S>0 and T>0, and where b>0 and b\neq 1, + +

                        + \log_bM=\log_bN if and only if M=N. +

                        + +

                        This is called the one-to-one property of logarithms

                        \log_bM=\log_bN if and only if M=N

                        From 24234346571b521d24fb49b83ec2e553ba619d86 Mon Sep 17 00:00:00 2001 From: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> Date: Wed, 18 Dec 2024 08:02:58 -0500 Subject: [PATCH 06/19] Update source/precalculus/source/05-EL/03.ptx --- source/precalculus/source/05-EL/03.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/03.ptx b/source/precalculus/source/05-EL/03.ptx index f1c5546d..ced39531 100644 --- a/source/precalculus/source/05-EL/03.ptx +++ b/source/precalculus/source/05-EL/03.ptx @@ -213,7 +213,7 @@                         - The powers of 10 function P(t) has an inverse L. This new function L is called the base 10 logarithm. But, we could have done a similar procedure with any base, which leads to the following definition. +

                        The powers of 10 function P(t) has an inverse L. This new function L is called the base 10 logarithm. But, we could have done a similar procedure with any base, which leads to the following definition.

                        From cfad711ec3d1280b084a5b8f45e55f1851cfff32 Mon Sep 17 00:00:00 2001 From: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> Date: Wed, 18 Dec 2024 08:04:54 -0500 Subject: [PATCH 07/19] added

                        tag to remark --- source/precalculus/source/05-EL/03.ptx | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/source/precalculus/source/05-EL/03.ptx b/source/precalculus/source/05-EL/03.ptx index ced39531..0b2231da 100644 --- a/source/precalculus/source/05-EL/03.ptx +++ b/source/precalculus/source/05-EL/03.ptx @@ -227,11 +227,11 @@ - We can use to express the relationship between logarithmic form and exponential form as follows: +

                        We can use to express the relationship between logarithmic form and exponential form as follows: \log_{b}(x)=y \iff b^{y}=x - whenever b \gt 0, b \neq 1 . + whenever b \gt 0, b \neq 1 .

                        From 1301a0ff4d7d661f5eab00c88eed42bf4c7f10e0 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:21:27 -0800 Subject: [PATCH 08/19] Update source/precalculus/source/05-EL/03.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/03.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/03.ptx b/source/precalculus/source/05-EL/03.ptx index 0b2231da..6c291fab 100644 --- a/source/precalculus/source/05-EL/03.ptx +++ b/source/precalculus/source/05-EL/03.ptx @@ -226,7 +226,7 @@

                        - +

                        We can use to express the relationship between logarithmic form and exponential form as follows: \log_{b}(x)=y \iff b^{y}=x From 483a4a6ddd5e96a7748bc914e41d770685952ee7 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:21:45 -0800 Subject: [PATCH 09/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 9318852b..16ab19ad 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -16,7 +16,7 @@ -

                        Recall from that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x.

                        +

                        Recall from that we can convert between exponential and logarithmic forms. \log_bx=y is equivalent to b^y=x.

                         From ac83ffcfe15b53f41555ae0d04a5583cb93bcbba Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:21:55 -0800 Subject: [PATCH 10/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 16ab19ad..f574f2b5 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -83,7 +83,7 @@                         - +

                        For any values S>0 and T>0, and where b>0 and b\neq 1, From 826fa73731f0547b1aa666acc1d93121d3d67dc8 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:22:06 -0800 Subject: [PATCH 11/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index f574f2b5..bd5a0aaf 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -98,7 +98,7 @@

                        -                         +

                        From b74e514eb968d9f2fea8e391c094b668f44aa8d1 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:22:28 -0800 Subject: [PATCH 12/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index bd5a0aaf..7051c9a6 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -102,7 +102,7 @@

                        - Notice this definition allows us to set the arguments (M and N) equal to each other once we get each side expressed as a single logarithm with matching bases. + Notice this fact will eventually help us solve logarithmic equations. If we have an equation where each side is expressed as a single logarithm with matching bases (such as \log_b M = \log_bN), then it follows that the arguments (M and N) are also equal to each other.

                        From 28ce3bf04c452c225e6ebadeab022c6694c84760 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:22:45 -0800 Subject: [PATCH 13/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 7051c9a6..96a0e60f 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -107,7 +107,8 @@ - Recall that exponentials and logarithms are inverses. We know that \log_b(b^{k})=k and according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} +

                        Recall that exponential functions and logarithmic functions are inverses. We know that \log_b(b^{k})=k and according to the law of exponents, we know that: + x^a \cdot x^b=x^{a+b} \dfrac{x^a}{x^b}=x^{a-b} \left(x^a\right)^b=x^{a \cdot b} From 91ed363f7522b6519d15ffe7ab9ad592cad5a9dc Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:23:07 -0800 Subject: [PATCH 14/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 96a0e60f..de7779e9 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -110,7 +110,7 @@

                        Recall that exponential functions and logarithmic functions are inverses. We know that \log_b(b^{k})=k and according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} \dfrac{x^a}{x^b}=x^{a-b} - \left(x^a\right)^b=x^{a \cdot b} + \left(x^a\right)^b=x^{a \cdot b}

                        Consider all these as you move through the next few activities in this section.                         From 35c827de9bf1b65f7d0debbce5bf0e55cbdd843e Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:23:16 -0800 Subject: [PATCH 15/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index de7779e9..2c9550f4 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -88,7 +88,7 @@

                        For any values S>0 and T>0, and where b>0 and b\neq 1, -

                        + \log_bM=\log_bN if and only if M=N.

                        From 58906ec24250d1f6b6f3961d8be0e95fcd73fc06 Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:23:29 -0800 Subject: [PATCH 16/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 2c9550f4..04a5efc0 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -92,7 +92,7 @@ \log_bM=\log_bN if and only if M=N.

                        -

                        This is called the one-to-one property of logarithms

                        +

                        This is called the one-to-one property of logarithms.

                        \log_bM=\log_bN if and only if M=N

                        From 23c41db5de5735022da458db3576ef6d8dcb791d Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:23:43 -0800 Subject: [PATCH 17/19] Update source/precalculus/source/05-EL/05.ptx Co-authored-by: Abby Noble <85640990+AbbyANoble@users.noreply.github.com> --- source/precalculus/source/05-EL/05.ptx | 4 ---- 1 file changed, 4 deletions(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index 04a5efc0..50dc86b4 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -93,10 +93,6 @@

                        This is called the one-to-one property of logarithms.

                        -

                        - \log_bM=\log_bN if and only if M=N -

                        -

                        From 8c65973ff69fc175263c39526dd23edf2342f62e Mon Sep 17 00:00:00 2001 From: Drew Lewis <30658947+siwelwerd@users.noreply.github.com> Date: Thu, 19 Dec 2024 15:24:36 -0800 Subject: [PATCH 18/19] Discard changes to source/precalculus/source/02-FN/05.ptx --- source/precalculus/source/02-FN/05.ptx | 132 ------------------------- 1 file changed, 132 deletions(-) diff --git a/source/precalculus/source/02-FN/05.ptx b/source/precalculus/source/02-FN/05.ptx index d47cec66..5418cc38 100644 --- a/source/precalculus/source/02-FN/05.ptx +++ b/source/precalculus/source/02-FN/05.ptx @@ -602,139 +602,7 @@ Let functions p and q be given by the graphs below.

                        - -

                        - Now that we've seen how to compose functions, let's take a look at how composition of functions can be applicable in our lives. -

                        -                         - - -

                        - Suppose you have a $50-off coupon and there is a 15\%-off sale on a TV. If you are allowed to apply both the coupon and sale price of the TV, which one should you apply first? Let's investigate to determine the better deal. -

                        -                         - - - -

                        - Write a cost function, C(p), for the price of the TV, p, if you applied the $50-off coupon. -

                        -                         - -

                        - C(p)=p-50 -

                        -                         -                         - - - -

                        - Write a cost function, S(p), for the price of the TV, p, if you applied the 15\%-off sale. -

                        -                         - -

                        - S(p)=p-.15p -

                        -

                        - S(p)=.85p -

                        -                         -                         - - - -

                        - Suppose you apply the coupon first and then the 15\%-off sale. Write a composite function to represent this situation. -

                        -                         - -

                        - S(C(p))=S(p-50) -

                        -

                        - S(C(p))=.85(p-50) -

                        -

                        - S(C(p))=.85p-42.5 -

                        -                         -                         - - -

                        - Suppose you apply the 15\%-off sale first and then the coupon. Write a composite function to represent this situation. -

                        -                         - -

                        - C(S(p))=C(.85p) -

                        -

                        - S(C(p))=(.85p)-50 -

                        -                         -                         - - - -

                        - Suppose the original cost of the TV is $500. Determine the cost of the TV if you applied the coupon first. -

                        -                         - -

                        - From part (c), we know that: S(C(p))=.85p-42.5 -

                        -

                        - S(C(500))=.85(500)-42.5 -

                        -

                        - S(C(500))=382.50 -

                        -

                        - When applying the coupon first, the reduced cost of the TV is $382.50. -

                        -                         -                         - - - -

                        - Suppose the original cost of the TV is $500. Determine the cost of the TV if you applied the 15\%-off sale first. -

                        -                         - -

                        - From part (d), we know that: S(C(p))=(.85p)-50 -

                        -

                        - S(C(500))=(.85(500))-50 -

                        -

                        - S(C(500))=375 -

                        -

                        - When applying the 15\%-off sale first, the reduced cost of the TV is $375. -

                        -                         -                         - - - -

                        - Which is the better deal and how much would you save? -

                        -                         - -

                        - When applying the coupon first, the TV would cost $382.50. If applying the 15\%-off sale first, however, the TV would cost $375. Applying the 15\%-off sale first is the better deal, saving you $125 (versus $117.50)! -

                        -                         -                         -                         From 1de414934083e07eb82c1c886b39e45d4d98eed8 Mon Sep 17 00:00:00 2001 From: Steven Clontz Date: Fri, 20 Dec 2024 02:32:37 +0000 Subject: [PATCH 19/19] fix mismatched p tag --- source/precalculus/source/05-EL/05.ptx | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/source/precalculus/source/05-EL/05.ptx b/source/precalculus/source/05-EL/05.ptx index c38202af..a0c41683 100644 --- a/source/precalculus/source/05-EL/05.ptx +++ b/source/precalculus/source/05-EL/05.ptx @@ -106,7 +106,7 @@

                        Recall that exponential functions and logarithmic functions are inverses. We know that \log_b(b^{k})=k and according to the law of exponents, we know that: x^a \cdot x^b=x^{a+b} \dfrac{x^a}{x^b}=x^{a-b} - \left(x^a\right)^b=x^{a \cdot b}

                        + \left(x^a\right)^b=x^{a \cdot b} Consider all these as you move through the activities in this section.