Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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~MRENTHUSIASM (Reconstruction) | ~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== | ==Solution 3== |
Revision as of 10:43, 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
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
Note that is the same as Using Reciprocal law, we get from which the answer is
~OlutosinNGA (Solution)
~MRENTHUSIASM (Reformatting)
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.