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

source/precalculus/source/02-FN/05.ptx

@@ -602,7 +602,139 @@ Let functions <m>p</m> and <m>q</m> be given by the graphs below. </p>
     </task>
 </activity>
 
+<remark>
+  <p>
+    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.
+  </p>
+</remark>
+<activity>
+  <introduction>
+    <p>
+      Suppose you have a <m>$50</m>-off coupon and there is a <m>15\%</m>-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.
+    </p>
+  </introduction>
+
+  <task>
+    <statement>
+      <p>
+        Write a cost function, <m>C(p)</m>, for the price of the TV, <m>p</m>, if you applied the <m>$50</m>-off coupon.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        <m>C(p)=p-50</m>
+      </p>
+    </answer>
+  </task>
+
+  <task>
+    <statement>
+      <p>
+        Write a cost function, <m>S(p)</m>, for the price of the TV, <m>p</m>, if you applied the <m>15\%</m>-off sale.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        <m>S(p)=p-.15p</m>
+      </p>
+      <p>
+        <m>S(p)=.85p</m>
+      </p>
+    </answer>
+  </task>
+
+  <task>
+    <statement>
+      <p>
+        Suppose you apply the coupon first and then the <m>15\%</m>-off sale. Write a composite function to represent this situation.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        <m>S(C(p))=S(p-50)</m>
+      </p>
+      <p>
+        <m>S(C(p))=.85(p-50)</m>
+      </p>
+      <p>
+        <m>S(C(p))=.85p-42.5</m>
+      </p>
+    </answer>
+  </task>
 
+  <task>
+    <statement>
+      <p>
+        Suppose you apply the <m>15\%</m>-off sale first and then the coupon. Write a composite function to represent this situation.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        <m>C(S(p))=C(.85p)</m>
+      </p>
+      <p>
+        <m>S(C(p))=(.85p)-50</m>
+      </p>
+    </answer>
+  </task>
+
+  <task>
+    <statement>
+      <p>
+        Suppose the original cost of the TV is <m>$500</m>. Determine the cost of the TV if you applied the coupon first.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        From part (c), we know that: <m>S(C(p))=.85p-42.5</m>
+      </p>
+      <p>
+        <m>S(C(500))=.85(500)-42.5</m> 
+      </p>
+      <p>
+        <m>S(C(500))=382.50</m> 
+      </p>
+      <p>
+        When applying the coupon first, the reduced cost of the TV is <m>$382.50</m>.
+      </p>
+    </answer>
+  </task>
+
+  <task>
+    <statement>
+      <p>
+        Suppose the original cost of the TV is <m>$500</m>. Determine the cost of the TV if you applied the <m>15\%</m>-off sale first.
+      </p>
+    </statement>
+    <answer>
+      <p>
+        From part (d), we know that: <m>S(C(p))=(.85p)-50</m>
+      </p>
+      <p>
+        <m>S(C(500))=(.85(500))-50</m>
+      </p>
+      <p>
+        <m>S(C(500))=375</m>
+      </p>
+      <p>
+        When applying the <m>15\%</m>-off sale first, the reduced cost of the TV is <m>$375</m>.
+      </p>
+    </answer>
+  </task>
+
+  <task>
+    <statement>
+      <p>
+        Which is the better deal and how much would you save?
+      </p>
+    </statement>
+    <answer>
+    <p>
+      When applying the coupon first, the TV would cost <m>$382.50</m>. If applying the <m>15\%</m>-off sale first, however, the TV would cost <m>$375</m>. Applying the <m>15\%</m>-off sale first is the better deal, saving you <m>$125</m> (versus <m>$117.50</m>)! 
+    </p>
+  </answer>
+  </task>
+</activity>
 
 
     </subsection>

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>
@@ -200,14 +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>
-  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.
+  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.
 </remark>
+
 <definition xml:id="def-log">
   <statement>
     <p>
@@ -223,7 +231,7 @@
   <me>
     \log_{b}(x)=y \iff b^{y}=x 
   </me>
-  whenever <m> b \gt 0, b \neq 1 </m>
+  whenever <m> b \gt 0, b \neq 1 </m>.
 </remark>
 <activity xml:id="act-log-exp">
   <introduction>

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

@@ -14,14 +14,104 @@
 <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="EL3"/> 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>
+
+    <definition 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>, 
+
+         <p>
+            <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>
+          <p>
+            <me>\log_bM=\log_bN</me> if and only if <me>M=N</me>
+          </p>
+        </p>
+      </statement>
+    </definition>
+
+    <observation>
+      <p>
+        Notice this definition allows us to set the arguments (<m>M</m> and <m>N</m>) equal to each other once we get each side expressed as a single logarithm with matching bases.
+      </p>
+    </observation>
 
   <remark>
-    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>
+    Recall that exponentials and logarithms 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> 
 
-      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.
    </remark>
 
    <activity xml:id="investigate-product-property-of-logs">
@@ -51,7 +141,7 @@
     <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>
@@ -89,7 +179,7 @@
     <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>
@@ -107,7 +197,7 @@
     <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>
@@ -123,7 +213,7 @@
     </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>
@@ -166,7 +256,7 @@
     <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>
@@ -204,7 +294,7 @@
     <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>
@@ -222,7 +312,7 @@
     <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>
@@ -238,7 +328,7 @@
     </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>