Difference between revisions of "2019 AIME II Problems/Problem 6"
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− | + | ==Problem== | |
+ | 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 | ||
+ | <cmath>3\log(\sqrt{x}\log x)=56</cmath> | ||
+ | <cmath>\log_{\log x}(x)=54</cmath> | ||
+ | and finds that this system of equations has a single real number solution <math>x>1</math>. Find <math>b</math>. | ||
+ | |||
+ | ==Solution 1== | ||
+ | Using change of base on the second equation to base b, | ||
+ | <cmath>\frac{\log x}{\log \log x }=54</cmath> | ||
+ | <cmath>\log x = 54 \cdot \log \log x</cmath> | ||
+ | <cmath>b^{\log x} = b^{54 \log \log x}</cmath> | ||
+ | <cmath>x = (b^{\log \log x})^{54}</cmath> | ||
+ | <cmath>x = (\log x)^{54}</cmath> | ||
+ | Note by dolphin7 - you could also just rewrite the second equation in exponent form. | ||
+ | Substituting this into the <math>\sqrt x</math> of the first equation, | ||
+ | <cmath>3\log((\log x)^{27}\log x) = 56</cmath> | ||
+ | <cmath>3\log(\log x)^{28} = 56</cmath> | ||
+ | <cmath>\log(\log x)^{84} = 56</cmath> | ||
+ | |||
+ | We can manipulate this equation to be able to substitute <math>x = (\log x)^{54}</math> a couple more times: | ||
+ | <cmath>\log(\log x)^{54} = 56 \cdot \frac{54}{84}</cmath> | ||
+ | <cmath>\log x = 36</cmath> | ||
+ | <cmath>(\log x)^{54} = 36^{54}</cmath> | ||
+ | <cmath>x = 6^{108}</cmath> | ||
+ | |||
+ | However, since we found that <math>\log x = 36</math>, <math>x</math> is also equal to <math>b^{36}</math>. Equating these, | ||
+ | <cmath>b^{36} = 6^{108}</cmath> | ||
+ | <cmath>b = 6^3 = \boxed{216}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We start by simplifying the first equation to | ||
+ | <cmath>3\log(\sqrt{x}\log x)=\log(x^{\frac{3}{2}}\log^3x)=56</cmath> | ||
+ | <cmath>x^\frac{3}{2}\cdot \log_b^3x=b^{56}</cmath> | ||
+ | Next, we simplify the second equation to | ||
+ | <cmath>\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54</cmath> | ||
+ | <cmath>\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))</cmath> | ||
+ | <cmath>x=\log_b^{54}x</cmath> | ||
+ | Substituting this into the first equation gives | ||
+ | <cmath>\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}</cmath> | ||
+ | <cmath>x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}</cmath> | ||
+ | Plugging this into <math>x=\log_b^{54}x</math> gives | ||
+ | <cmath>b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}</cmath> | ||
+ | <cmath>b^{\frac{2}{3}}=36</cmath> | ||
+ | <cmath>b=36^{\frac{3}{2}}=6^3=\boxed{216}</cmath> | ||
+ | -ktong | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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> | ||
+ | which can be rearranged as: <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> | ||
+ | Apply log properties to <cmath>3\log(\sqrt{x}\log x)=56</cmath> to yield: | ||
+ | <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> | ||
+ | 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> | ||
+ | So <cmath>\log_b(x)=36.</cmath> | ||
+ | Substituting this back in to <cmath>\frac{\log_b(x)}{54}=\log_b(\log_b(x))</cmath> yields | ||
+ | <cmath>\frac{36}{54}=\log_b(36).</cmath> | ||
+ | So, | ||
+ | <cmath>b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}</cmath> | ||
+ | |||
+ | -Ghazt2002 | ||
+ | |||
+ | ==Solution 4== | ||
+ | 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>. | ||
+ | |||
+ | -Stormersyle | ||
+ | |||
+ | ==Solution 5 (Substitution)== | ||
+ | Let <math>y = \log _{b} x</math> | ||
+ | Then we have | ||
+ | <cmath>3\log _{b} (y\sqrt{x}) = 56</cmath> | ||
+ | <cmath>\log _{y} x = 54</cmath> | ||
+ | which gives | ||
+ | <cmath>y^{54} = x</cmath> | ||
+ | Plugging this in gives | ||
+ | <cmath>3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56</cmath> | ||
+ | which gives | ||
+ | <cmath>\log _{b} y = \dfrac{2}{3}</cmath> | ||
+ | so | ||
+ | <cmath>b^{2/3} = y</cmath> | ||
+ | By substitution we have | ||
+ | <cmath>b^{36} = x</cmath> | ||
+ | which gives | ||
+ | <cmath>y = \log _{b} x = 36</cmath> | ||
+ | Plugging in again we get | ||
+ | <cmath>b = 36^{3/2} = \fbox{216}</cmath> | ||
+ | |||
+ | --Hi3142 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|num-b=5|num-a=7}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 14:01, 18 April 2021
Contents
Problem
In a Martian civilization, all logarithms whose bases are not specified as assumed to be base , for some fixed . A Martian student writes down and finds that this system of equations has a single real number solution . Find .
Solution 1
Using change of base on the second equation to base b, Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the of the first equation,
We can manipulate this equation to be able to substitute a couple more times:
However, since we found that , is also equal to . Equating these,
Solution 2
We start by simplifying the first equation to Next, we simplify the second equation to Substituting this into the first equation gives Plugging this into gives -ktong
Solution 3
Apply change of base to to yield: which can be rearranged as: Apply log properties to to yield: Substituting into the equation yields: So Substituting this back in to yields So,
-Ghazt2002
Solution 4
1st equation: 2nd equation: So now substitute and : We also have that This means that , so .
-Stormersyle
Solution 5 (Substitution)
Let Then we have which gives Plugging this in gives which gives so By substitution we have which gives Plugging in again we get
--Hi3142
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
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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