Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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− | == Problem == | + | ==Problem== |
The solutions to the equation <math>\log_{3x} 4 = \log_{2x} 8</math>, where <math>x</math> is a positive real number other than <math>\tfrac{1}{3}</math> or <math>\tfrac{1}{2}</math>, can be written as <math>\tfrac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q</math>? | The solutions to the equation <math>\log_{3x} 4 = \log_{2x} 8</math>, where <math>x</math> is a positive real number other than <math>\tfrac{1}{3}</math> or <math>\tfrac{1}{2}</math>, can be written as <math>\tfrac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q</math>? | ||
Line 9: | Line 9: | ||
\textbf{(E) } 35 </math> | \textbf{(E) } 35 </math> | ||
− | == Solution == | + | ==Solution 1== |
+ | We apply the Change of Base Formula, then rearrange: | ||
+ | <cmath>\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*}</cmath> | ||
+ | By the logarithmic identity <math>n\log_b{a}=\log_b{\left(a^n\right)},</math> it follows that | ||
+ | <cmath>\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*}</cmath> | ||
+ | from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math> | ||
+ | ~jeremylu (Fundamental Logic) | ||
− | + | ~MRENTHUSIASM (Reconstruction) | |
− | < | + | ==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>\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: | ||
+ | <cmath>\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*}</cmath> | ||
+ | Therefore, the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math> | ||
+ | ~Pikachu13307 (Fundamental Logic) | ||
+ | |||
+ | ~MRENTHUSIASM (Reconstruction) | ||
+ | |||
+ | ==Solution 4== | ||
+ | We can convert both <math>4</math> and <math>8</math> into <math>2^2</math> and <math>2^3,</math> respectively: <cmath>2\log_{3x} (2) = 3\log_{2x} (2).</cmath> | ||
+ | Converting the bases of the right side, we get | ||
+ | <cmath>\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*}</cmath> | ||
+ | Dividing both sides by <math>\ln 2,</math> we get <math>\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)},</math> from which <cmath>2\ln (2x) = 3\ln (3x).</cmath> | ||
+ | Expanding this equation gives | ||
+ | <cmath>\begin{align*} | ||
+ | 2\ln 2 + 2\ln (x) &= 3\ln 3 + 3\ln (x) \\ | ||
+ | \ln (x) &= 2\ln 2 - 3\ln 3. | ||
+ | \end{align*}</cmath> | ||
+ | Thus, we have <cmath>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},</cmath> | ||
+ | from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math> | ||
+ | |||
+ | ~lepetitmoulin (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Reformatting) | ||
+ | |||
+ | ==Solution 5 (Exponential Form)== | ||
+ | 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*} | ||
+ | (3x)^y&=4, \\ | ||
+ | (2x)^y&=8. | ||
+ | \end{align*}</cmath> | ||
+ | Cubing the first equation and squaring the second equation, we have | ||
+ | <cmath>\begin{align*} | ||
+ | (3x)^{3y}&=64, \\ | ||
+ | (2x)^{2y}&=64. | ||
+ | \end{align*}</cmath> | ||
+ | Applying the Transitive Property, we get | ||
+ | <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> | ||
+ | |||
+ | ~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
Contents
Problem
The solutions to the equation , where is a positive real number other than or , can be written as where and are relatively prime positive integers. What is ?
Solution 1
We apply the Change of Base Formula, then rearrange: By the logarithmic identity it follows that from which the answer is
~jeremylu (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
We will apply the following logarithmic identity: which can be proven by the Change of Base Formula: We rewrite the original equation as from which Therefore, the answer is
~MRENTHUSIASM
Solution 3
By the logarithmic identity the original equation becomes By the logarithmic identity we multiply both sides by then apply the Change of Base Formula to the left side: Therefore, the answer is
~Pikachu13307 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 4
We can convert both and into and respectively: Converting the bases of the right side, we get Dividing both sides by we get from which Expanding this equation gives Thus, we have from which the answer is
~lepetitmoulin (Solution)
~MRENTHUSIASM (Reformatting)
Solution 5 (Exponential Form)
Let We convert the equations with to the exponential form: Cubing the first equation and squaring the second equation, we have Applying the Transitive Property, we get from which the answer is
~MRENTHUSIASM
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
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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