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fix typo and clean up EL5 activity #946

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siwelwerd merged 1 commit into from
Mar 26, 2026
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fix typo and clean up EL5 activity #946

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65 changes: 15 additions & 50 deletions source/precalculus/source/05-EL/05.ptx
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Original file line number Diff line number Diff line change
Expand Up @@ -236,25 +236,8 @@
<task>
<statement>
<p>
Let <m>a=10^x</m> and <m>b=10^y</m>. How could you rewrite the left side of the equation <m>\dfrac{10^x}{10^y}</m>?
<ol marker= "A." cols="2">
<li><m>a+b</m></li>
<li><m>a-b</m></li>
<li><m>10^{x+y}</m></li>
<li><m>a \cdot b</m></li></ol>
</p>
</statement>
<answer>
<p>
B
</p>
</answer>
</task>

<task>
<statement>
<p>
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?
If we let <m>a=10^x</m> and <m>b=10^y</m> and apply a logarithm
to both sides of the equation, what would be the result?
<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>
Expand All @@ -276,10 +259,10 @@
Recall that this property was covered in <xref ref="EL3"/>.
</aside>
<ol marker= "A." cols="2">
<li><m>\log_{10}(a-b)</m></li>
<li><m>\log_{10}(x-y)</m></li>
<li><m>x-y</m></li>
<li><m>a-b</m></li></ol>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=\log_{10}(a-b)</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=\log_{10}(x-y)</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=x-y</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=a-b</m></li></ol>
</p>
</statement>
<answer>
Expand All @@ -292,12 +275,12 @@
<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? (Choose two.)
<ol marker= "A." cols="2">
<li><m>\log_{10}a=x</m></li>
<li><m>\log_{x}a=10</m></li>
<li><m>\log_{10}b=y</m></li>
<li><m>\log_{y}b=10</m></li></ol>
<li><m>x=\log_{10}a</m></li>
<li><m>x=\log_{a}10</m></li>
<li><m>y=\log_{10}b</m></li>
<li><m>y=\log_{b}10</m></li></ol>
</p>
</statement>
<answer>
Expand All @@ -306,32 +289,14 @@
</p>
</answer>
</task>

<task>
<statement>
<p>
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>
<li><m>\log_{10}a-\log_{10}b</m></li>
<li><m>10^{x-y}</m></li></ol>
</p>
</statement>
<answer>
<p>
B
</p>
</answer>
</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 these results, 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>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=\log_{10}a+\log_{10}b</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=10^{x-y}</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=\log_{10}a-\log_{10}b</m></li>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=10^{x-y}</m></li></ol>
<li><m>\log_{10}\left(\dfrac{a}{b}\right)=10^{x+y}</m></li></ol>
</p>
</statement>
<answer>
Expand Down
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