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1964 AHSME Problems/Problem 21 - Revision history
2024-03-28T11:27:18Z
Revision history for this page on the wiki
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https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_21&diff=107860&oldid=prev
Talkinaway: /* Solution 2 */
2019-07-24T06:45:10Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 06:45, 24 July 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l52" >Line 52:</td>
<td colspan="2" class="diff-lineno">Line 52:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the definition of a logarithm, the first term on the left is asking for the exponent <math>x</math> needed to change the number <math>b^2</math> into <math>(b^2)^x</math> to get to <math>b^n</math>.  That exponent is <math>\frac{n}{2}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>By the definition of a logarithm, the first term on the left is asking for the exponent <math>x</math> needed to change the number <math>b^2</math> into <math>(b^2)^x</math> to get to <math>b^n</math>.  That exponent is <math>\frac{n}{2}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The second term is asking for a similar exponent needed to change <math>b^{2n}</math> into <math>b</math>.  That exponent is <del class="diffchange diffchange-inline">also </del><math>\frac{1}{2n}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The second term is asking for a similar exponent needed to change <math>b^{2n}</math> into <math>b</math>.  That exponent is <math>\frac{1}{2n}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The equation becomes <math>\frac{n}{2} + \frac{1}{2n} = 1</math>.  Multiplying by <math>2n</math> gives the quadratic <math>n^2 + 1 = 2n</math>, which has the solution <math>n=1</math>.  Thus, <math>x = b^n = b^1</math>, and the answer is <math>\boxed{\textbf{(D)}}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>The equation becomes <math>\frac{n}{2} + \frac{1}{2n} = 1</math>.  Multiplying by <math>2n</math> gives the quadratic <math>n^2 + 1 = 2n</math>, which has the solution <math>n=1</math>.  Thus, <math>x = b^n = b^1</math>, and the answer is <math>\boxed{\textbf{(D)}}</math>.</div></td></tr>
</table>
Talkinaway
https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_21&diff=107859&oldid=prev
Talkinaway at 06:44, 24 July 2019
2019-07-24T06:44:07Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 06:44, 24 July 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\textbf{(E)}\ \sqrt{b} </math>     </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\textbf{(E)}\ \sqrt{b} </math>     </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== Solution==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== Solution <ins class="diffchange diffchange-inline">1</ins>==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Using natural log as a "neutral base", and applying the change of base formula to each term, we get:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Using natural log as a "neutral base", and applying the change of base formula to each term, we get:</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l41" >Line 41:</td>
<td colspan="2" class="diff-lineno">Line 41:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Either way, the answer is <math>\boxed{\textbf{(D)}}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Either way, the answer is <math>\boxed{\textbf{(D)}}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">==Solution 2==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">All answers are of the form <math>x = b^n</math>, so we substitute that into the equation and try to solve for <math>n</math>.  We get:</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>\log_{b^2}x+\log_{x^2}b=1</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>\log_{b^2}b^n + \log_{b^{2n}} b = 1</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">By the definition of a logarithm, the first term on the left is asking for the exponent <math>x</math> needed to change the number <math>b^2</math> into <math>(b^2)^x</math> to get to <math>b^n</math>.  That exponent is <math>\frac{n}{2}</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The second term is asking for a similar exponent needed to change <math>b^{2n}</math> into <math>b</math>.  That exponent is also <math>\frac{1}{2n}</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The equation becomes <math>\frac{n}{2} + \frac{1}{2n} = 1</math>.  Multiplying by <math>2n</math> gives the quadratic <math>n^2 + 1 = 2n</math>, which has the solution <math>n=1</math>.  Thus, <math>x = b^n = b^1</math>, and the answer is <math>\boxed{\textbf{(D)}}</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See Also==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==See Also==</div></td></tr>
</table>
Talkinaway
https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_21&diff=107856&oldid=prev
Talkinaway: /* Solution */
2019-07-24T02:42:16Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-content" />
<col class="diff-marker" />
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:42, 24 July 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l26" >Line 26:</td>
<td colspan="2" class="diff-lineno">Line 26:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>You could inspect the equation here and see that <math>x=b</math> is one solution.  Or, you can substitute <math>X = \ln x</math> and <math>B = \ln b</math> to get a quadratic in <math>X</math>:</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>You could inspect the equation here and see that <math>x=b</math> is one solution.  Or, you can substitute <math>X = \ln x</math> and <math>B = \ln b</math> to get a quadratic in <math>X</math>:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>X^2 <del class="diffchange diffchange-inline">- </del>B^2 = 2BX</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>X^2 <ins class="diffchange diffchange-inline">+ </ins>B^2 = 2BX</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><math>X^2 - 2BX <del class="diffchange diffchange-inline">- </del>B^2 = 0</math></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><math>X^2 - 2BX <ins class="diffchange diffchange-inline">+ </ins>B^2 = 0</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>The above is a quadratic with coefficients <math>(1, -2B, <del class="diffchange diffchange-inline">-</del>B^2)</math>.  Plug into the QF to get:</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>The above is a quadratic with coefficients <math>(1, -2B, B^2)</math>.  Plug into the QF to get:</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}</math></div></td></tr>
</table>
Talkinaway
https://artofproblemsolving.com/wiki/index.php?title=1964_AHSME_Problems/Problem_21&diff=107855&oldid=prev
Talkinaway: Created page with "== Problem 21== If <math>\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1</math>, then <math>x</math> equals: <math>\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \te..."
2019-07-24T02:39:23Z
<p>Created page with "== Problem 21== If <math>\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1</math>, then <math>x</math> equals: <math>\textbf{(A)}\ 1/b^2 \qquad \textbf{(B)}\ 1/b \qquad \te..."</p>
<p><b>New page</b></p><div>== Problem 21==<br />
<br />
If <math>\log_{b^2}x+\log_{x^2}b=1, b>0, b \neq 1, x \neq 1</math>, then <math>x</math> equals:<br />
<br />
<math>\textbf{(A)}\ 1/b^2 \qquad<br />
\textbf{(B)}\ 1/b \qquad<br />
\textbf{(C)}\ b^2 \qquad<br />
\textbf{(D)}\ b \qquad<br />
\textbf{(E)}\ \sqrt{b} </math> <br />
<br />
== Solution==<br />
<br />
Using natural log as a "neutral base", and applying the change of base formula to each term, we get:<br />
<br />
<math>\frac{\ln x}{\ln b^2} + \frac{\ln b}{\ln x^2} = 1</math><br />
<br />
<br />
<math>\frac{\ln x}{2\ln b} + \frac{\ln b}{2\ln x} = 1</math><br />
<br />
<br />
<math>\frac{\ln x \ln x + \ln b \ln b}{2\ln b \ln x} = 1</math><br />
<br />
<br />
<math>\ln x \ln x + \ln b \ln b = 2\ln b \ln x</math><br />
<br />
You could inspect the equation here and see that <math>x=b</math> is one solution. Or, you can substitute <math>X = \ln x</math> and <math>B = \ln b</math> to get a quadratic in <math>X</math>:<br />
<br />
<math>X^2 - B^2 = 2BX</math><br />
<br />
<math>X^2 - 2BX - B^2 = 0</math><br />
<br />
The above is a quadratic with coefficients <math>(1, -2B, -B^2)</math>. Plug into the QF to get:<br />
<br />
<math>X = \frac{2B \pm \sqrt{4B^2 - 4B^2}}{2}</math><br />
<br />
<math>X = B</math><br />
<br />
<math>\ln x = \ln b</math><br />
<br />
<math>x = b</math><br />
<br />
Either way, the answer is <math>\boxed{\textbf{(D)}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1964|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
<br />
{{MAA Notice}}</div>
Talkinaway