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

(See Also)
(Solution)
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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>
 
Set y to log x. Then the equations become:
 
3log(<math>\sqrt x</math>y)=54, and
 
<cmath>\log_{\y}(x)=54</cmath>
 
  
 
==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 19:06, 22 March 2019

Problem 6

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

Using change of base on the second equation to base b, \[\frac{\log x}{\log \log x }=54\] \[\log x = 54 \cdot \log \log x\] \[b^{\log x} = b^{54 \log \log x}\] \[x = (b^{\log \log x})^{54}\] \[x = (\log x)^{54}\] 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, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\]

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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