Difference between revisions of "2019 AIME II Problems/Problem 6"
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+ | ==Problem 6== | ||
+ | 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== | ||
+ | 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> | ||
+ | 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> | ||
+ | |||
+ | Using <math>x = (\log x)^{54}</math> again, | ||
+ | <cmath>\frac{84}{54}\log(\log x)^{54} = 56</cmath> | ||
+ | <cmath>\frac{14}{9}\log x = 56</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> |
Revision as of 17:46, 22 March 2019
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
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,
Substituting this into the
of the first equation,
Using again,
However, since we found that ,
is also equal to
. Equating these,