Difference between revisions of "2019 AIME II Problems/Problem 6"
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− | ==Problem | + | ==Problem== |
− | In a Martian civilization, all logarithms whose bases are not specified | + | In a Martian civilization, all logarithms whose bases are not specified are 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>3\log(\sqrt{x}\log x)=56</cmath> | ||
<cmath>\log_{\log x}(x)=54</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>. | and finds that this system of equations has a single real number solution <math>x>1</math>. Find <math>b</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Using change of base on the second equation to base b, | Using change of base on the second equation to base b, | ||
− | <cmath>\frac{\log x}{\log \ | + | <cmath>\frac{\log x}{\log \log_{b}{x}}=54</cmath> |
− | <cmath>\log x = 54 \cdot \log \ | + | <cmath>\log x = 54 \cdot \log\log_{b}{x}</cmath> |
− | + | <cmath>x = (\log_{b}{x})^{54}</cmath> | |
− | <cmath>x = (b | + | 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, | Substituting this into the <math>\sqrt x</math> of the first equation, | ||
− | <cmath>3\ | + | <cmath>3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56</cmath> |
− | <cmath>3\ | + | <cmath>3\log_{b}{(\log_{b}{x})^{28}} = 56</cmath> |
− | <cmath>\ | + | <cmath>\log_{b}{(\log_{b}{x})^{84}} = 56</cmath> |
− | + | We can manipulate this equation to be able to substitute <math>x = (\log_{b}{x})^{54}</math> a couple more times: | |
− | <cmath>\ | + | <cmath>\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}</cmath> |
− | + | <cmath>\log_{b}{x} = 36</cmath> | |
− | <cmath>\ | + | <cmath>(\log_{b}{x})^{54} = 36^{54}</cmath> |
− | <cmath>(\ | ||
<cmath>x = 6^{108}</cmath> | <cmath>x = 6^{108}</cmath> | ||
− | However, since we found that <math>\ | + | However, since we found that <math>\log_{b}{x} = 36</math>, <math>x</math> is also equal to <math>b^{36}</math>. Equating these, |
<cmath>b^{36} = 6^{108}</cmath> | <cmath>b^{36} = 6^{108}</cmath> | ||
<cmath>b = 6^3 = \boxed{216}</cmath> | <cmath>b = 6^3 = \boxed{216}</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We start by simplifying the first equation to | ||
+ | <cmath>3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(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 | ||
+ | |||
+ | ==Solution 6 (Also Substitution)== | ||
+ | This system of equations looks complicated to work with, so we let <math>a=\log_bx</math> to make it easier for us to read. | ||
+ | |||
+ | Now, the first equation becomes <math>3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3</math>. | ||
+ | |||
+ | The second equation, <math>\log_{\log(x)}(x)=54</math> gives us <math>\underline{a^{54} = x}</math>. | ||
+ | |||
+ | Let's plug this back into the first equation to see what we get: <math>\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3</math>, and simplifying, <math>\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}</math>, so <math>b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}</math>. | ||
+ | |||
+ | Combining this new finding with what we had above <math>a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}</math>. | ||
+ | |||
+ | Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get <math>\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies </math><math>\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36</math>. | ||
+ | |||
+ | Finally, that gives us that <math>\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3</math>. Thus, <math>b=\boxed{216}</math>. | ||
+ | |||
+ | ~BakedPotato66 | ||
+ | |||
+ | ==Solution 7 (Easy System of Equations)== | ||
+ | Using change of base on the second equation, we have | ||
+ | |||
+ | <cmath>\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54</cmath> | ||
+ | |||
+ | Using log rules on the first equation, we have | ||
+ | |||
+ | <cmath>\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56</cmath> | ||
+ | |||
+ | We notice that <math>\log_{b} x</math> and <math>\log_{b} \log_{b} x</math> are in both equations. Thus, we set <math>m = \log_{b} x</math> and <math>n = \log_{b} \log_{b} x</math> and we have | ||
+ | |||
+ | <cmath>\frac{3}{2} m + 3n = 56</cmath> | ||
+ | <cmath>\frac{m}{n} = 54</cmath> | ||
+ | |||
+ | Solving this yields <math>m = 36</math>, <math>n = \frac{2}{3}</math>. | ||
+ | |||
+ | Now, <math>n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36</math>, so we have <math>\log_{b} 36 = \frac{2}{3}</math>. Solving this yields <math>b = \boxed{216}</math>. | ||
+ | |||
+ | ~ adam_zheng | ||
+ | |||
+ | ==Solution 8 (Definition of Logarithm)== | ||
+ | The second equation implies that | ||
+ | <cmath>\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}</cmath> | ||
+ | The first equation implies that | ||
+ | <cmath>3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x</cmath> | ||
+ | Substituting the first result into the second gives us | ||
+ | <cmath>b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.</cmath> | ||
+ | Because <math>b^{36}=x,</math> <math>\log_b x=36</math> by the definition of a logarithm. | ||
+ | Substituting this into the second equation, | ||
+ | <cmath>\log_{36} x=54 \implies x=36^{54}.</cmath> | ||
+ | Finally, | ||
+ | <cmath>b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.</cmath> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2019|n=II|num-b=5|num-a=7}} | ||
+ | [[Category: Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 09:09, 5 April 2024
Contents
Problem
In a Martian civilization, all logarithms whose bases are not specified are 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
Solution 6 (Also Substitution)
This system of equations looks complicated to work with, so we let to make it easier for us to read.
Now, the first equation becomes .
The second equation, gives us .
Let's plug this back into the first equation to see what we get: , and simplifying, , so .
Combining this new finding with what we had above .
Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get .
Finally, that gives us that . Thus, .
~BakedPotato66
Solution 7 (Easy System of Equations)
Using change of base on the second equation, we have
Using log rules on the first equation, we have
We notice that and are in both equations. Thus, we set and and we have
Solving this yields , .
Now, , so we have . Solving this yields .
~ adam_zheng
Solution 8 (Definition of Logarithm)
The second equation implies that The first equation implies that Substituting the first result into the second gives us Because by the definition of a logarithm. Substituting this into the second equation, Finally,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.