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

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==Problem==
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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 1==
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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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Note by dolphin7 - you could also just rewrite the second equation in exponent form.
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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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We can manipulate this equation to be able to substitute <math>x = (\log x)^{54}</math> a couple more times:
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<cmath>\log(\log x)^{54} = 56 \cdot \frac{54}{84}</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>
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==Solution 2==
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We start by simplifying the first equation to
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<cmath>3\log(\sqrt{x}\log x)=\log(x^{\frac{3}{2}}\log^3x)=56</cmath>
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<cmath>x^\frac{3}{2}\cdot \log_b^3x=b^{56}</cmath>
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Next, we simplify the second equation to
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<cmath>\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54</cmath>
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<cmath>\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))</cmath>
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<cmath>x=\log_b^{54}x</cmath>
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Substituting this into the first equation gives
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<cmath>\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}</cmath>
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<cmath>x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}</cmath>
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Plugging this into <math>x=\log_b^{54}x</math> gives
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<cmath>b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}</cmath>
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<cmath>b^{\frac{2}{3}}=36</cmath>
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<cmath>b=36^{\frac{3}{2}}=6^3=\boxed{216}</cmath>
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-ktong
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==Solution 3==
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Apply change of base to <cmath>\log_{\log x}(x)=54</cmath> to yield: <cmath>\frac{\log_b(x)}{\log_b(\log_b(x))}=54</cmath>
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which can be rearranged as:  <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath>
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Apply log properties to <cmath>3\log(\sqrt{x}\log x)=56</cmath> to yield:
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<cmath>3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}</cmath>
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Substituting <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> into the equation <math>\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}</math> yields: <cmath>\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}</cmath>
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So <cmath>\log_b(x)=36.</cmath>
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Substituting this back in to <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> yields
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<cmath>\frac{36}{54}=\log_b(36).</cmath>
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So,
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<cmath>b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}</cmath>
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-Ghazt2002
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==Solution 4==
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1st equation: <cmath>\log (\sqrt{x}\log x)=\frac{56}{3}</cmath> <cmath>\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}</cmath> 2nd equation: <cmath>x=(\log x)^{54}</cmath> So now substitute <math>\log x=a</math> and <math>x=b^a</math>: <cmath>b^a=a^{54}</cmath> <cmath>b=a^{\frac{54}{a}}</cmath> We also have that <cmath>\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}</cmath> <cmath>\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}</cmath> This means that <math>\frac{14}{27}a=\frac{56}{3}</math>, so <cmath>a=36</cmath> <cmath>b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}</cmath>.
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-Stormersyle
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==Solution 5 (Substitution)==
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Let <math>y = \log _{b} x</math>
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Then we have
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<cmath>3\log _{b} (y\sqrt{x}) = 56</cmath>
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<cmath>\log _{y} x = 54</cmath>
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which gives
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<cmath>y^{54} = x</cmath>
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Plugging this in gives
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<cmath>3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56</cmath>
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which gives
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<cmath>\log _{b} y = \dfrac{2}{3}</cmath>
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so
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<cmath>b^{2/3} = y</cmath>
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By substitution we have
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<cmath>b^{36} = x</cmath>
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which gives
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<cmath>y = \log _{b} x = 36</cmath>
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Plugging in again we get
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<cmath>b = 36^{3/2} = \fbox{216}</cmath>
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--Hi3142
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==See Also==
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{{AIME box|year=2019|n=II|num-b=5|num-a=7}}
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 14:01, 18 April 2021

Problem

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 1

Using change of base on the second equation to base b, \[\frac{\log x}{\log \log x }=54\] \[\log x = 54 \cdot \log \log x\] \[b^{\log x} = b^{54 \log \log x}\] \[x = (b^{\log \log x})^{54}\] \[x = (\log x)^{54}\] Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the $\sqrt x$ of the first equation, \[3\log((\log x)^{27}\log x) = 56\] \[3\log(\log x)^{28} = 56\] \[\log(\log x)^{84} = 56\]

We can manipulate this equation to be able to substitute $x = (\log x)^{54}$ a couple more times: \[\log(\log x)^{54} = 56 \cdot \frac{54}{84}\] \[\log x = 36\] \[(\log x)^{54} = 36^{54}\] \[x = 6^{108}\]

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

Solution 2

We start by simplifying the first equation to \[3\log(\sqrt{x}\log x)=\log(x^{\frac{3}{2}}\log^3x)=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] Next, we simplify the second equation to \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\log_b^{54}x\] Substituting this into the first equation gives \[\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}\] \[x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}\] Plugging this into $x=\log_b^{54}x$ gives \[b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}\] \[b^{\frac{2}{3}}=36\] \[b=36^{\frac{3}{2}}=6^3=\boxed{216}\] -ktong

Solution 3

Apply change of base to \[\log_{\log x}(x)=54\] to yield: \[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] which can be rearranged as: \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] Apply log properties to \[3\log(\sqrt{x}\log x)=56\] to yield: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\] Substituting \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: \[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\] So \[\log_b(x)=36.\] Substituting this back in to \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] yields \[\frac{36}{54}=\log_b(36).\] So, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}\]

-Ghazt2002

Solution 4

1st equation: \[\log (\sqrt{x}\log x)=\frac{56}{3}\] \[\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}\] 2nd equation: \[x=(\log x)^{54}\] So now substitute $\log x=a$ and $x=b^a$: \[b^a=a^{54}\] \[b=a^{\frac{54}{a}}\] We also have that \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}\] This means that $\frac{14}{27}a=\frac{56}{3}$, so \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\].

-Stormersyle

Solution 5 (Substitution)

Let $y = \log _{b} x$ Then we have \[3\log _{b} (y\sqrt{x}) = 56\] \[\log _{y} x = 54\] which gives \[y^{54} = x\] Plugging this in gives \[3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56\] which gives \[\log _{b} y = \dfrac{2}{3}\] so \[b^{2/3} = y\] By substitution we have \[b^{36} = x\] which gives \[y = \log _{b} x = 36\] Plugging in again we get \[b = 36^{3/2} = \fbox{216}\]

--Hi3142

See Also

2019 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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