Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 5(Substitution))
(→‎Solution 4 (Easiest))
Line 59: Line 59:
  
 
==Solution 4 (Easiest)==
 
==Solution 4 (Easiest)==
−
From the first equation we have that <math>\log (\sqrt{x}\log x)=\frac{56}{3}</math>, so <math>\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}</math>. From the second equation we have that <math>x=(\log x)^{54}</math>, so now set <math>\log x=a</math> and <math>x=b^a</math>. Substituting, we have that <math>b^a=a^{54}</math>, so <math>b=a^{\frac{54}{a}}</math>. We also have that <math>\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}</math>, so <math>\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}</math>. This means that <math>\frac{14}{27}a=\frac{56}{3}</math>, so <math>a=36</math>, and <math>b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}</math>.
+
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
 
-Stormersyle

Revision as of 02:29, 5 November 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 1

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, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\]

Solution 2

We start by simplifying the first equation to \[3\log(\sqrt{x}\log x)=\log(x^{\frac{3}{2}}\log^3x)=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] Next, we simplify the second equation to \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\log_b^{54}x\] Substituting this into the first equation gives \[\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}\] \[x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}\] Plugging this into $x=\log_b^{54}x$ gives \[b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}\] \[b^{\frac{2}{3}}=36\] \[b=36^{\frac{3}{2}}=6^3=\boxed{216}\] -ktong

Solution 3 (Elegant)

Apply change of base to \[\log_{\log x}(x)=54\] to yield: \[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] which can be rearranged as: \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] Apply log properties to \[3\log(\sqrt{x}\log x)=56\] to yield: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\] Substituting \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: \[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\] So \[\log_b(x)=36.\] Substituting this back in to \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] yields \[\frac{36}{54}=\log_b(36).\] So, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{b=216}\]

-Ghazt2002

Solution 4 (Easiest)

1st equation: \[\log (\sqrt{x}\log x)=\frac{56}{3}\] \[\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}\] 2nd equation: \[x=(\log x)^{54}\] So now substitute $\log x=a$ and $x=b^a$: \[b^a=a^{54}\] \[b=a^{\frac{54}{a}}\] We also have that \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}\] This means that $\frac{14}{27}a=\frac{56}{3}$, so \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\].

-Stormersyle

Solution 5(Substitution)

Let $y = \log _{b} x$ Then we have \[3\log _{b} (y\sqrt{x}) = 56\] \[\log _{y} x = 54\] which gives \[y^{54} = x\] Plugging this in gives \[3\log _{b} (y * y^{27}) = 3\log _{b} y^{28} = 56\] which gives \[\log _{b} y = \dfrac{2}{3}\] so \[b^{2/3} = y\] By substitution we have \[b^{36} = x\] which gives \[y = \log _{b} x = 36\] Plugging in again we get \[b = 36^{3/2} = \fbox{216}\]

--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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_6&oldid=110904"