Difference between revisions of "2018 AMC 12A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Reformatted Sol 4. Will insert Sol 3 later.) |
MRENTHUSIASM (talk | contribs) m (Polished Sol 5.) |
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==Solution 5== | ==Solution 5== | ||
− | <math>\log_{3x} 4=\log_{2x} 8</math> is the same as < | + | Note that <math>\log_{3x} 4=\log_{2x} 8</math> is the same as <cmath>2\log_{3x} 2=3\log_{2x} 2.</cmath> |
+ | Using Reciprocal law, we get | ||
+ | <cmath>\begin{align*} | ||
+ | \log_{(3x)^\frac{1}{2}} 2&=\log_{(2x)^\frac{1}{3}} 2 \\ | ||
+ | (3x)^\frac{1}{2}&=(2x)^\frac{1}{3} \\ | ||
+ | 27x^3&=4x^2 \\ | ||
+ | x&=\frac{4}{27}, | ||
+ | \end{align*}</cmath> | ||
+ | from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math> | ||
− | + | ~OlutosinNGA (Solution) | |
− | + | ~MRENTHUSIASM (Reformatting) | |
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− | |||
− | |||
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==Solution 6== | ==Solution 6== |
Revision as of 09:18, 14 August 2021
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
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
Note that is the same as Using Reciprocal law, we get from which the answer is
~OlutosinNGA (Solution)
~MRENTHUSIASM (Reformatting)
Solution 6
We know that Thus and are indeed relatively prime thus our final answer is
-vsamc
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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