Difference between revisions of "2018 AMC 12A Problems/Problem 14"
MRENTHUSIASM (talk | contribs) (Sol 6 is quite similar to the solutions above, so I decide to delete it. Let me know if anyone is unhappy about it ...) |
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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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~MRENTHUSIASM (Reconstruction) | ~MRENTHUSIASM (Reconstruction) | ||
− | ==Solution | + | ==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}. |
Revision as of 10:37, 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 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.