Difference between revisions of "2019 AIME II Problems/Problem 6"
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<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== | ||
| + | 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 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|num-b=5|num-a=7}} | {{AIME box|year=2019|n=II|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:41, 22 March 2019
Contents
Problem 6
In a Martian civilization, all logarithms whose bases are not specified as assumed to be base 
, for some fixed 
. A Martian student writes down
![\[3\log(\sqrt{x}\log x)=56\]](http://latex.artofproblemsolving.com/e/d/d/edd14880fb8a07eb5e444758008df636655d25fa.png)
![\[\log_{\log x}(x)=54\]](http://latex.artofproblemsolving.com/a/4/f/a4f7d9de7f66eef8b9752b307cdda8a029dbc98b.png)
and finds that this system of equations has a single real number solution 
. Find .
Solution
Using change of base on the second equation to base b,
![\[\frac{\log x}{\log \log x }=54\]](http://latex.artofproblemsolving.com/c/5/a/c5a0acad77288f08cd18d7f064427de108da5076.png)
![\[\log x = 54 \cdot \log \log x\]](http://latex.artofproblemsolving.com/7/2/9/729214250a0f1c280de2bbe48df67ce5ca461ab9.png)
![\[b^{\log x} = b^{54 \log \log x}\]](http://latex.artofproblemsolving.com/2/5/e/25e1e20d02f37431dffaecaeab8c811f960e2979.png)
![\[x = (b^{\log \log x})^{54}\]](http://latex.artofproblemsolving.com/e/7/8/e781be47a659b925c7cb69d9e9a729943fc2242d.png)
![\[x = (\log x)^{54}\]](http://latex.artofproblemsolving.com/2/8/0/280f73065f9aa89dbd6c31c0ecdbe388c6b8f014.png)
Substituting this into the 
of the first equation,
![\[3\log((\log x)^{27}\log x) = 56\]](http://latex.artofproblemsolving.com/e/f/5/ef52ec0742eb072dbaef0dcb79a0a0731d7a7b4e.png)
![\[3\log(\log x)^{28} = 56\]](http://latex.artofproblemsolving.com/1/7/6/176148de48a598ef03474ab62d8c1266c92b1c88.png)
![\[\log(\log x)^{84} = 56\]](http://latex.artofproblemsolving.com/1/0/3/1036924d8ffa6173ec964bf099997d1fa2587223.png)
We can manipulate this equation to be able to substitute 
a couple more times:
![\[\log(\log x)^{54} = 56 \cdot \frac{54}{84}\]](http://latex.artofproblemsolving.com/9/2/7/927d7448063e4ae392e4f88b276a19d2280b5d73.png)
![\[\log x = 36\]](http://latex.artofproblemsolving.com/1/7/7/177eda721fc80448e7453e2726c5de27f6fdfad4.png)
![\[(\log x)^{54} = 36^{54}\]](http://latex.artofproblemsolving.com/7/0/5/70501ddaae7ede4ebe0effb8200c083c6c7420fe.png)
![\[x = 6^{108}\]](http://latex.artofproblemsolving.com/6/2/7/627365b85149d5158f48cbfb0044770173ab5016.png)
However, since we found that 
, 
is also equal to 
. Equating these,
![\[b^{36} = 6^{108}\]](http://latex.artofproblemsolving.com/5/2/3/523bdb1d74bfee78c672bb44d6f739ca59ec62f1.png)
![\[b = 6^3 = \boxed{216}\]](http://latex.artofproblemsolving.com/4/f/c/4fcc283ea20c219448778290bd3b586d051ea06f.png)
Solution 2
Apply change of base to to yield: ![\[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\]](http://latex.artofproblemsolving.com/6/a/0/6a0bdd0b431c65a028309a4d20c752c670d6ceaf.png)
which can be rearranged as: ![\[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\]](http://latex.artofproblemsolving.com/c/b/6/cb68e6c66d8d82479a35dd7bade5275c581fd836.png)
Apply log properties to to yield:
![\[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\]](http://latex.artofproblemsolving.com/3/2/e/32ed7121c40f0d3d6f47798fa1b272f5f7ede371.png)
Substituting into the equation 
yields: ![\[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\]](http://latex.artofproblemsolving.com/5/d/1/5d1f0ef646894df91a4e242576f094cd7b3b842e.png)
So ![\[\log_b(x)=36.\]](http://latex.artofproblemsolving.com/2/e/c/2ec1df14c0202e21f4a33cc755c594956db7143a.png)
Substituting this back in to yields
![\[\frac{36}{54}=\log_b(36).\]](http://latex.artofproblemsolving.com/f/0/7/f073afb5622d44b79019813bbe0731a5497b4bee.png)
So,
![\[b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}\]](http://latex.artofproblemsolving.com/3/7/e/37ee1a897cff8515ac6839db87bf3b56649aaf3d.png)
-Ghazt2002
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. 
