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Difference between revisions of "2019 AIME II Problems/Problem 6"

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==Problem 6==
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In a Martian civilization, all logarithms whose bases are not specified as assumed to be base <math>b</math>, for some fixed <math>b\ge2</math>. A Martian student writes down
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<cmath>3\log(\sqrt{x}\log x)=56</cmath>
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<cmath>\log_{\log x}(x)=54</cmath>
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and finds that this system of equations has a single real number solution <math>x>1</math>. Find <math>b</math>.
  
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==Solution==
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Using change of base on the second equation to base b,
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<cmath>\frac{\log x}{\log \log x }=54</cmath>
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<cmath>\log x = 54 \cdot \log \log x</cmath>
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<cmath>b^{\log x} = b^{54 \log \log x}</cmath>
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<cmath>x = (b^{\log \log x})^{54}</cmath>
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<cmath>x = (\log x)^{54}</cmath>
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Substituting this into the <math>\sqrt x</math> of the first equation,
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<cmath>3\log((\log x)^{27}\log x) = 56</cmath>
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<cmath>3\log(\log x)^{28} = 56</cmath>
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<cmath>\log(\log x)^{84} = 56</cmath>
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Using <math>x = (\log x)^{54}</math> again,
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<cmath>\frac{84}{54}\log(\log x)^{54} = 56</cmath>
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<cmath>\frac{14}{9}\log x = 56</cmath>
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<cmath>\log x = 36</cmath>
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<cmath>(\log x)^{54} = 36^{54}</cmath>
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<cmath>x = 6^{108}</cmath>
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However, since we found that <math>\log x = 36</math>, <math>x</math> is also equal to <math>b^{36}</math>. Equating these,
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<cmath>b^{36} = 6^{108}</cmath>
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<cmath>b = 6^3 = \boxed{216}</cmath>

Revision as of 16:46, 22 March 2019

Problem 6

In a Martian civilization, all logarithms whose bases are not specified as assumed to be base $b$, for some fixed $b\ge2$. A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$. Find $b$.

Solution

Using change of base on the second equation to base b, \[\frac{\log x}{\log \log x }=54\] \[\log x = 54 \cdot \log \log x\] \[b^{\log x} = b^{54 \log \log x}\] \[x = (b^{\log \log x})^{54}\] \[x = (\log x)^{54}\] Substituting this into the $\sqrt x$ of the first equation, \[3\log((\log x)^{27}\log x) = 56\] \[3\log(\log x)^{28} = 56\] \[\log(\log x)^{84} = 56\]

Using $x = (\log x)^{54}$ again, \[\frac{84}{54}\log(\log x)^{54} = 56\] \[\frac{14}{9}\log x = 56\] \[\log x = 36\] \[(\log x)^{54} = 36^{54}\] \[x = 6^{108}\]

However, since we found that $\log x = 36$, $x$ is also equal to $b^{36}$. Equating these, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\]

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