Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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==Solution 5 (Exponential Form)== | ==Solution 5 (Exponential Form)== | ||
− | Let <math>y=\log_{3x} 4 = \log_{2x} 8.</math> We convert | + | Let <math>y=\log_{3x} 4 = \log_{2x} 8.</math> We convert each equation for <math>y</math> to the exponential form: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
(3x)^y&=4, \\ | (3x)^y&=4, \\ |
Revision as of 20:36, 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 each equation for 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.