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"

(→‎Solution)
(→‎Solution 7 (Easy System of Equations))
 
(27 intermediate revisions by 15 users not shown)
Line 1: Line 1:
−
==Problem 6==
+
==Problem==
−
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
+
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 \log x }=54</cmath>
+
<cmath>\frac{\log x}{\log \log_{b}{x}}=54</cmath>
−
<cmath>\log x = 54 \cdot \log \log x</cmath>
+
<cmath>\log x = 54 \cdot \log\log_{b}{x}</cmath>
−
<cmath>b^{\log x} = b^{54 \log \log x}</cmath>
+
<cmath>x = (\log_{b}{x})^{54}</cmath>
−
<cmath>x = (b^{\log \log x})^{54}</cmath>
+
Note by dolphin7 - you could also just rewrite the second equation in exponent form.
−
<cmath>x = (\log x)^{54}</cmath>
 
 
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\log((\log x)^{27}\log x) = 56</cmath>
+
<cmath>3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56</cmath>
−
<cmath>3\log(\log x)^{28} = 56</cmath>
+
<cmath>3\log_{b}{(\log_{b}{x})^{28}} = 56</cmath>
−
<cmath>\log(\log x)^{84} = 56</cmath>
+
<cmath>\log_{b}{(\log_{b}{x})^{84}} = 56</cmath>
  
−
We can manipulate this equation to be able to substitute <math>x = (\log x)^{54}</math> a couple more times:
+
We can manipulate this equation to be able to substitute <math>x = (\log_{b}{x})^{54}</math> a couple more times:
−
<cmath>\log(\log x)^{54} = 56 \cdot \frac{54}{84}</cmath>
+
<cmath>\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}</cmath>
−
<cmath>\log x = 36</cmath>
+
<cmath>\log_{b}{x} = 36</cmath>
−
<cmath>(\log x)^{54} = 36^{54}</cmath>
+
<cmath>(\log_{b}{x})^{54} = 36^{54}</cmath>
 
<cmath>x = 6^{108}</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,
+
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==
 
==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}}
 +
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:09, 5 April 2024

Problem

In a Martian civilization, all logarithms whose bases are not specified are 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, \[\frac{\log x}{\log \log_{b}{x}}=54\] \[\log x = 54 \cdot \log\log_{b}{x}\] \[x = (\log_{b}{x})^{54}\] Note by dolphin7 - you could also just rewrite the second equation in exponent form. Substituting this into the $\sqrt x$ of the first equation, \[3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56\] \[3\log_{b}{(\log_{b}{x})^{28}} = 56\] \[\log_{b}{(\log_{b}{x})^{84}} = 56\]

We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times: \[\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}\] \[\log_{b}{x} = 36\] \[(\log_{b}{x})^{54} = 36^{54}\] \[x = 6^{108}\]

However, since we found that $\log_{b}{x} = 36$, $x$ 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 \[3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] 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

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

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 \[3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56\] 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

Solution 6 (Also Substitution)

This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.

Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$.

The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$.

Let's plug this back into the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$, and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$, so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$.

Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$.

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 $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$$\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$.

Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$. Thus, $b=\boxed{216}$.

~BakedPotato66

Solution 7 (Easy System of Equations)

Using change of base on the second equation, we have

\[\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54\]

Using log rules on the first equation, we have

\[\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56\]

We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = \log_{b} \log_{b} x$ and we have

\[\frac{3}{2} m + 3n = 56\] \[\frac{m}{n} = 54\]

Solving this yields $m = 36$, $n = \frac{2}{3}$.

Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$, so we have $\log_{b} 36 = \frac{2}{3}$. Solving this yields $b = \boxed{216}$.

~ adam_zheng

Solution 8 (Definition of Logarithm)

The second equation implies that \[\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}\] The first equation implies that \[3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x\] Substituting the first result into the second gives us \[b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.\] Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm. Substituting this into the second equation, \[\log_{36} x=54 \implies x=36^{54}.\] Finally, \[b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.\]

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=217854"