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?