Fixed One-to-one property of logs in EL3, EL5, and EL6 #430

source/precalculus/source/05-EL/03.ptx

@@ -109,7 +109,7 @@
   <task>
     <statement>
       <p>
-        Since <m>P</m> has an inverse function, we know there exists some other function, say <m>L</m>, such that <m>y=P(t)</m> represent the same relationship between <m>t</m> and <m>y</m> as <m>t=L(y)</m>.  In words, this means that <m>L</m> reverses the process of raising to the power of 10, telling us the <em>power</em> to which we need to raise 10 to produce a desired result.
+        Since <m>P</m> has an inverse function, we know there exists some other function, say <m>L</m>, such that <m>y=P(t)</m> represent the same relationship between <m>t</m> and <m>y</m> as <m>t=L(y)</m>.  In words, this means that <m>L</m> reverses the process of raising to the power of <m>10</m>, telling us the <em>power</em> to which we need to raise <m>10</m> to produce a desired result.
 
         Fill in the table of values for <m>L(y)</m>.
         <tabular halign="right">
@@ -188,7 +188,10 @@
     </statement>
     <answer>
       <p>
-        Domain <m>(-\infty, \infty)</m> Range <m>(0,\infty)</m>
+        Domain: <m>(-\infty, \infty)</m>
+      </p>
+      <p>
+        Range: <m>(0,\infty)</m>
       </p>
     </answer>
   </task>
@@ -200,16 +203,19 @@
       </statement>
       <answer>
         <p>
-          Domain <m>(0,\infty)</m> Range <m>(-\infty, \infty)</m>
+          Domain: <m>(0,\infty)</m>
+        </p>
+        <p>
+          Range: <m>(-\infty, \infty)</m>
         </p>
       </answer>
   </task>
 </activity>
+
 <remark>
-<p>
-  The powers of 10 function <m>P(t)</m> has an inverse <m>L</m>.  This new function <m>L</m> is called the base 10 logarithm.  But, we could have done a similar procedure with any base, which leads to the following definition.
-</p>
+<p>  The powers of <m>10</m> function <m>P(t)</m> has an inverse <m>L</m>.  This new function <m>L</m> is called the base <m>10</m> logarithm.  But, we could have done a similar procedure with any base, which leads to the following definition. </p>
 </remark>
+
 <definition xml:id="def-log">
   <statement>
     <p>
@@ -220,14 +226,12 @@
     </p>
   </statement>
 </definition>
-<remark>
-<p>
-  We can use <xref ref="def-log"/> to express the relationship between logarithmic form and exponential form as follows:
+<remark xml:id="remark-log-exp-form">
+<p>  We can use <xref ref="def-log"/> to express the relationship between logarithmic form and exponential form as follows:
   <me>
     \log_{b}(x)=y \iff b^{y}=x 
   </me>
-  whenever <m> b \gt 0, b \neq 1 </m>
-</p>
+  whenever <m> b \gt 0, b \neq 1 </m>.</p>
 </remark>
 <activity xml:id="act-log-exp">
   <introduction>

source/precalculus/source/05-EL/05.ptx

@@ -14,13 +14,99 @@
 <introduction>
   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.
   </introduction>
+
+  <remark>
+     <p> Recall from <xref ref="remark-log-exp-form"/> that we can convert between exponential and logarithmic forms. <me>\log_bx=y</me> is equivalent to <me>b^y=x</me>.</p>
+    </remark>
+
+    <activity xml:id="intro-to-one-to-one-property">
+    <introduction>
+      <p>
+        Suppose you are given two equations
+        <m>\log_bM=x</m> and <m>\log_bN=x</m>.
+      </p>
+    </introduction>
+
+    <task>
+      <statement>
+        <p>
+          Rewrite each logarithmic equation into an exponential equation.
+        </p>
+      </statement>
+      <answer>
+        <p>
+          <m>\log_bM=x \iff b^{x}=M </m>
+        </p>
+        <p>
+          <m>\log_bN=x \iff b^{x}=N </m>
+        </p>
+      </answer>
+    </task>
+
+    <task>
+      <statement>
+        <p>
+          Look at the exponential equations you found in part (a). What conclusion can you make about <m>M</m> and <m>N</m>?
+        </p>
+      </statement>
+      <answer>
+        <p>
+          Because <m>b^{x}=M</m> and <m>b^{x}=N</m>, then it follows that <m>M=N</m>.
+        </p>
+      </answer>
+    </task>
+
+    <task>
+      <statement>
+        <p>
+          Given that both <m>\log_bM</m> and <m>\log_bN</m> are both equal to <m>x</m>, what can you say about <m>\log_bM</m> and <m>\log_bN</m>?
+        </p>
+      </statement>
+      <answer>
+        <p>
+          <m>\log_bM=\log_bN</m>
+        </p>
+      </answer>
+    </task>
+
+    <task>
+      <statement>
+        <p>
+          If given <m>\log_bM=\log_bN</m>, what can you say about <m>M</m> and <m>N</m>? (Refer back to the previous parts of this activity.)
+        </p>
+      </statement>
+      <answer>
+        <p>
+          Based on what we have seen in parts (b) and (c), we can conclude that <m>M=N</m>.
+        </p>
+      </answer>
+    </task>
+    </activity>
+
+    <fact xml:id="def-one-to-one-property">
+      <statement>
+        <p>
+         For any values <m>S>0</m> and <m>T>0</m>,  and where <m>b>0</m> and <m>b\neq 1</m>, 
+
+
+            <me>\log_bM=\log_bN</me> if and only if <me>M=N.</me>
+          </p>
+
+        <p> This is called the <term>one-to-one property of logarithms</term>.</p>
+      </statement>
+    </fact>
+
+    <observation>
+      <p>
+        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 <m>\log_b M = \log_bN</m>), then it follows that the arguments (<m>M</m> and <m>N</m>) are also equal to each other.
+      </p>
+    </observation>
 
   <remark>
-  <p>
-    Recall that <m>\log_b M=\log_b N</m> if and only if <m>M=N</m>. In addition, because exponentials and logarithms are inverses, we also know that <m>\log_b(b^{k})=k</m>.
-    In addition, according to the law of exponents, we know that: <me>x^a \cdot x^b=x^{a+b}</me>
+ <p>   Recall that exponential functions and logarithmic functions are inverses. We know that <m>\log_b(b^{k})=k</m> and according to the law of exponents, we know that: 
+ <me>x^a \cdot x^b=x^{a+b}</me>
       <me>\dfrac{x^a}{x^b}=x^{a-b}</me>
-      <me>\left(x^a\right)^b=x^{a \cdot b}</me> 
+      <me>\left(x^a\right)^b=x^{a \cdot b}</me>
 
       Consider all these as you move through the activities in this section.
     </p>
@@ -53,7 +139,7 @@
     <task>
       <statement>
         <p>
-          Recall from <xref ref="EL3"/> that <m>\log_b M=\log_b N</m> if and only if <m>M=N</m>. Use this property to apply the logarithm to both sides of the rewritten equation from part a. What is that equation?
+          Recall from <xref ref="def-one-to-one-property"/> that <m>\log_b M=\log_b N</m> if and only if <m>M=N</m>. Use this property to apply the logarithm to both sides of the rewritten equation from part (a). What is that equation?
           <ol marker= "A." cols="1">
             <li><m>\log_{10}(a+b)=\log_{10}\left(10^{x+y}\right)</m></li>
             <li><m>\log_{10}(a \cdot b)=\log_{10}\left(10^{x+y}\right)</m></li>
@@ -91,7 +177,7 @@
     <task>
       <statement>
         <p>
-          Recall in part a, we defined <m>10^x=a</m> and <m>10^y=b</m>. What would these look like in logarithmic form?
+          Recall in part (a), we defined <m>10^x=a</m> and <m>10^y=b</m>. What would these look like in logarithmic form?
           <ol marker= "A." cols="2">
             <li><m>\log_{10}a=x</m></li>
             <li><m>\log_{x}a=10</m></li>
@@ -109,7 +195,7 @@
     <task>
       <statement>
         <p>
-          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?
           <ol marker= "A." cols="1">
             <li><m>10^{a+b}</m></li>
             <li><m>\log_{10}a-\log_{10}b</m></li>
@@ -125,7 +211,7 @@
     </task>
     <task>
       <statement>
-        <p> Combining parts a and d, which equation represents <m>10^x \cdot 10^y=10^{x+y}</m> in terms of logarithms?
+        <p> Combining parts (a) and (d), which equation represents <m>10^x \cdot 10^y=10^{x+y}</m> in terms of logarithms?
           <ol marker= "A." cols="1">
             <li><m>\log_{10}(a+b)=10^{a+b}</m></li>
             <li><m>\log_{10}(a \cdot b)=\log_{10}a-\log_{10}b</m></li>
@@ -168,7 +254,7 @@
     <task>
       <statement>
         <p>
-          Recall from <xref ref="EL3"/> that <m>\log_b M=\log_b N</m> if and only if <m>M=N</m>. 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?
           <ol marker= "A." cols="1">
             <li><m>\log_{10}(a-b)=\log_{10}\left(10^{x-y}\right)</m></li>
             <li><m>\log_{10}\left(\dfrac{a}{b}\right)=\log_{10}\left(10^{x-y}\right)</m></li>
@@ -206,7 +292,7 @@
     <task>
       <statement>
         <p>
-          Recall in part a, we defined <m>10^x=a</m> and <m>10^y=b</m>. What would these look like in logarithmic form?
+          Recall in part (a), we defined <m>10^x=a</m> and <m>10^y=b</m>. What would these look like in logarithmic form?
           <ol marker= "A." cols="2">
             <li><m>\log_{10}a=x</m></li>
             <li><m>\log_{x}a=10</m></li>
@@ -224,7 +310,7 @@
     <task>
       <statement>
         <p>
-          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?
           <ol marker= "A." cols="1">
             <li><m>10^{a+b}</m></li>
             <li><m>\log_{10}a-\log_{10}b</m></li>
@@ -240,7 +326,7 @@
     </task>
     <task>
       <statement>
-        <p> Combining parts a and d, which equation represents <m>\dfrac{10^x}{10^y}=10^{x-y}</m> in terms of logarithms?
+        <p> Combining parts (a) and (d), which equation represents <m>\dfrac{10^x}{10^y}=10^{x-y}</m> in terms of logarithms?
           <ol marker= "A." cols="1">
             <li><m>\log_{10}(a-b)=10^{a+b}</m></li>
             <li><m>\log_{10}(a-b)=\log_{10}a-\log_{10}b</m></li>

source/precalculus/source/05-EL/06.ptx

@@ -169,7 +169,7 @@
       <task>
         <statement>
           <p>
-            Notice that the "log" has disappeared and you now have an equation with just the variable <m>x</m>. 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 <m>x</m>. Which of the following is equivalent to the equation you got in part (b)?
             <ol marker= "A." cols="1">
               <li><m>4x-5=2x-1</m></li>
               <li><m>4x-5=0</m></li>
@@ -185,7 +185,7 @@
       <task>
         <statement>
           <p>
-            Compare the answer you got in part c to the original equation given <m>\log(4x-5)=\log(2x-1)</m>. What do you notice?
+            Compare the answer you got in part (c) to the original equation given <m>\log(4x-5)=\log(2x-1)</m>. What do you notice?
           </p>
         </statement>
         <answer>
@@ -197,7 +197,7 @@
       <task>
         <statement>
           <p>
-            Solve the equation you got in part d to find the value of <m>x</m>.
+            Solve the equation you got in part (d) to find the value of <m>x</m>.
             <ol marker= "A." cols="2">
               <li><m>-3</m></li>
               <li><m>\dfrac{5}{4}</m></li>
@@ -219,18 +219,10 @@
     </p>
     </remark> 
 
-    <definition xml:id="def-one-to-one-property">
-      <statement>
-        <p>
-          The <term>one-to-one property of logarithms</term> 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).
-        </p>
-      </statement>
-    </definition>
-
     <activity xml:id="solving-using-one-to-one-property">
       <introduction>
         <p>
-          Apply <xref ref="def-one-to-one-property"/> 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 <xref ref="def-one-to-one-property"/>) and other properties of logarithms (i.e., product, quotient, and power) to solve the following logarithmic equations.
         </p>
       </introduction>
       <task>
@@ -357,7 +349,7 @@
     <task>
       <statement>
         <p>
-          Notice that the answer you got in part b is an exact answer for <m>x</m>. There will be times, though, that it will be helpful to also have an approximation for <m>x</m>. Which of the following is a good approximation for <m>x</m>?
+          Notice that the answer you got in part (b) is an exact answer for <m>x</m>. There will be times, though, that it will be helpful to also have an approximation for <m>x</m>. Which of the following is a good approximation for <m>x</m>?
           <ol marker= "A." cols="2">
             <li><m>x \approx 5.585</m></li>
             <li><m>x \approx 0.179</m></li>
@@ -424,7 +416,7 @@
       <task>
         <statement>
           <p>
-            Using the change-of-base formula (<xref ref="def-change-of-base"/>), rewrite your answer from part c so that <m>x</m> is written as a single logarithm. What is the exact value of <m>x</m>?
+            Using the change-of-base formula (<xref ref="def-change-of-base"/>), rewrite your answer from part (c) so that <m>x</m> is written as a single logarithm. What is the exact value of <m>x</m>?
           </p>
         </statement>
         <answer>
@@ -448,12 +440,12 @@
       <task>
         <statement>
           <p>
-            What do you notice about your answer from parts d and e?
+            What do you notice about your answer from parts (d) and (e)?
           </p>
         </statement>
         <answer>
           <p>
-            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.
           </p>
         </answer>
       </task>
@@ -580,7 +572,7 @@
       <task>
         <statement>
           <p>
-            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?
           </p>
         </statement>
         <answer>