Difference between revisions of "2018 AMC 12A Problems/Problem 14"

(Reformatted Sol 4. Will insert Sol 3 later.)
m (Solution 5 (Exponential Form))
 
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\
 
\log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\
(3x)^3&=(2x)^2\\
+
(3x)^3&=(2x)^2 \\
 
27x^3&=4x^2 \\
 
27x^3&=4x^2 \\
 
x&=\frac{4}{27},
 
x&=\frac{4}{27},
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==Solution 2==
 
==Solution 2==
 +
We will apply the following logarithmic identity:
 +
<cmath>\log_{p^n}{\left(q^n\right)}=\log_{p}{q},</cmath>
 +
which can be proven by the Change of Base Formula: <cmath>\log_{p^n}{\left(q^n\right)}=\frac{\log_{p}{\left(q^n\right)}}{\log_{p}{\left(p^n\right)}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.</cmath>
 +
We rewrite the original equation as <math>\log_{(3x)^3} 64 = \log_{(2x)^2} 64,</math> from which
 +
<cmath>\begin{align*}
 +
(3x)^3&=(2x)^2 \\
 +
27x^3&=4x^2 \\
 +
x&=\frac{4}{27}.
 +
\end{align*}</cmath>
 +
Therefore, the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 3==
 
By the logarithmic identity <math>n\log_b{a}=\log_b{\left(a^n\right)},</math> the original equation becomes <cmath>2\log_{3x} 2 = 3\log_{2x} 2.</cmath>
 
By the logarithmic identity <math>n\log_b{a}=\log_b{\left(a^n\right)},</math> the original equation becomes <cmath>2\log_{3x} 2 = 3\log_{2x} 2.</cmath>
 
By the logarithmic identity <math>\log_b{a}\cdot\log_a{b}=1,</math> we multiply both sides by <math>\log_2{(2x)},</math> then apply the Change of Base Formula to the left side:
 
By the logarithmic identity <math>\log_b{a}\cdot\log_a{b}=1,</math> we multiply both sides by <math>\log_2{(2x)},</math> then apply the Change of Base Formula to the left side:
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2\left[\log_{3x}{(2x)}\right] &= 3 \\
 
2\left[\log_{3x}{(2x)}\right] &= 3 \\
 
\log_{3x}{\left[(2x)^2\right]} &= 3 \\
 
\log_{3x}{\left[(2x)^2\right]} &= 3 \\
(3x)^3&=(2x)^2\\
+
(3x)^3&=(2x)^2 \\
 
27x^3&=4x^2 \\
 
27x^3&=4x^2 \\
 
x&=\frac{4}{27}.
 
x&=\frac{4}{27}.
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~MRENTHUSIASM (Reformatting)
 
~MRENTHUSIASM (Reformatting)
  
==Solution 5==
+
==Solution 5 (Exponential Form)==
<math>\log_{3x} 4=\log_{2x} 8</math> is the same as <math>2\log_{3x} 2=3\log_{2x} 2</math>
+
Let <math>y=\log_{3x} 4 = \log_{2x} 8.</math> We convert the equations with <math>y</math> to the exponential form:
 
+
<cmath>\begin{align*}
Using Reciprocal law, we get <math>\log_{(3x)^\frac{1}{2}} 2=\log_{(2x)^\frac{1}{3}} 2</math>
+
(3x)^y&=4, \\
 
+
(2x)^y&=8.
<math>\Rightarrow (3x)^\frac{1}{2}=(2x)^\frac{1}{3}</math> <math>\Rightarrow 27x^3=4x^2</math> <math>\Rightarrow \frac{x^3}{x^2}=\frac{4}{27}=x</math>
+
\end{align*}</cmath>
 
+
Cubing the first equation and squaring the second equation, we have
<math>\therefore \frac{p}{q}=\frac{4}{27}</math> <math>\Rightarrow p+q=4+27=</math> <math>\boxed{\textbf{(D) } 31}</math>
+
<cmath>\begin{align*}
 
+
(3x)^{3y}&=64, \\
~OlutosinNGA
+
(2x)^{2y}&=64.
 
+
\end{align*}</cmath>
==Solution 6==
+
Applying the Transitive Property, we get
<math>\log_{3x} 4=\log_{2x} 8\implies 2\log_{3x} 2=3\log_{2x} 2 \implies \frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}.</math> We know that <math>\log_a{b}=\frac{\log_{c}b}{\log_{c}a}=\frac{\frac{1}{\log_b{c}}}{\frac{1}{\log_a{c}}}=\frac{\log_a{c}}{\log_b{c}}.</math> Thus <math>\frac{2}{3}=\frac{\log_{2x}2}{\log_{2x}3}\implies \frac{2}{3}=\log_{2x}{3x}\implies (2x)^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}x^{\frac{2}{3}}=3x\implies 2^{\frac{2}{3}}=3x^{\frac{1}{3}}\implies x^{\frac{1}{3}}=\frac{2^{\frac{2}{3}}}{3}\implies x=\frac{2^2}{3^3}=\frac{4}{27}.</math> <math>4</math> and <math>27</math> are indeed relatively prime thus our final answer is <math>4+27=31 \text{which is }\boxed{\textbf{(D)}}</math>
+
<cmath>\begin{align*}
 +
(3x)^{3y}&=(2x)^{2y} \\
 +
(3x)^3&=(2x)^2 \\
 +
27x^3&=4x^2 \\
 +
x&=\frac{4}{27},
 +
\end{align*}</cmath>
 +
from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math>
  
-vsamc
+
~MRENTHUSIASM
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:32, 15 August 2021

Problem

The solutions to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$?

$\textbf{(A) } 5   \qquad     \textbf{(B) } 13   \qquad     \textbf{(C) } 17   \qquad    \textbf{(D) } 31 \qquad   \textbf{(E) } 35$

Solution 1

We apply the Change of Base Formula, then rearrange: \begin{align*} \frac{\log_2{4}}{\log_2{(3x)}}&=\frac{\log_2{8}}{\log_2{(2x)}} \\ \frac{2}{\log_2{(3x)}}&=\frac{3}{\log_2{(2x)}} \\ 3\log_2{(3x)}&=2\log_2{(2x)}. \\ \end{align*} By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ it follows that \begin{align*} \log_2{\left[(3x)^3\right]}&=\log_2{\left[(2x)^2\right]} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~jeremylu (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 2

We will apply the following logarithmic identity: \[\log_{p^n}{\left(q^n\right)}=\log_{p}{q},\] which can be proven by the Change of Base Formula: \[\log_{p^n}{\left(q^n\right)}=\frac{\log_{p}{\left(q^n\right)}}{\log_{p}{\left(p^n\right)}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.\] We rewrite the original equation as $\log_{(3x)^3} 64 = \log_{(2x)^2} 64,$ from which \begin{align*} (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~MRENTHUSIASM

Solution 3

By the logarithmic identity $n\log_b{a}=\log_b{\left(a^n\right)},$ the original equation becomes \[2\log_{3x} 2 = 3\log_{2x} 2.\] By the logarithmic identity $\log_b{a}\cdot\log_a{b}=1,$ we multiply both sides by $\log_2{(2x)},$ then apply the Change of Base Formula to the left side: \begin{align*} 2\left[\log_{3x}2\right]\left[\log_2{(2x)}\right] &= 3 \\ 2\left[\frac{\log_2 2}{\log_2{(3x)}}\right]\left[\frac{\log_2{(2x)}}{\log_2 2}\right] &= 3 \\ 2\left[\frac{\log_2{(2x)}}{\log_2{(3x)}}\right] &=3 \\ 2\left[\log_{3x}{(2x)}\right] &= 3 \\ \log_{3x}{\left[(2x)^2\right]} &= 3 \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}. \end{align*} Therefore, the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~Pikachu13307 (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 4

We can convert both $4$ and $8$ into $2^2$ and $2^3,$ respectively: \[2\log_{3x} (2) = 3\log_{2x} (2).\] Converting the bases of the right side, we get \begin{align*} \log_{2x} 2 &= \frac{\ln 2}{\ln (2x)} \\ \frac{2}{3}\cdot\log_{3x} (2) &= \frac{\ln 2}{\ln (2x)} \\ 2^\frac{2}{3} &= (3x)^\frac{\ln 2}{\ln (2x)} \\ \frac{2}{3} \cdot \ln 2 &= \frac{\ln 2}{\ln (2x)} \cdot \ln (3x). \end{align*} Dividing both sides by $\ln 2,$ we get $\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},$ from which \[2\ln (2x) = 3\ln (3x).\] Expanding this equation gives \begin{align*} 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ \ln (x) &= 2\ln 2 - 3\ln 3. \end{align*} Thus, we have \[x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}} = \frac{2^2}{3^3} = \frac{4}{27},\] from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~lepetitmoulin (Solution)

~MRENTHUSIASM (Reformatting)

Solution 5 (Exponential Form)

Let $y=\log_{3x} 4 = \log_{2x} 8.$ We convert the equations with $y$ to the exponential form: \begin{align*} (3x)^y&=4, \\ (2x)^y&=8. \end{align*} Cubing the first equation and squaring the second equation, we have \begin{align*} (3x)^{3y}&=64, \\ (2x)^{2y}&=64. \end{align*} Applying the Transitive Property, we get \begin{align*} (3x)^{3y}&=(2x)^{2y} \\ (3x)^3&=(2x)^2 \\ 27x^3&=4x^2 \\ x&=\frac{4}{27}, \end{align*} from which the answer is $4+27=\boxed{\textbf{(D) } 31}.$

~MRENTHUSIASM

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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