Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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==Solution 1== | ==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 | ||
+ | <math></math>\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*} | ||
+ | from which the answer is <math>4+27=\boxed{\textbf{(D) } 31}.</math> | ||
+ | ~jeremylu (Fundamental Logic) | ||
− | + | ~MRENTHUSIASM (Reconstruction) | |
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==Solution 2== | ==Solution 2== |
Revision as of 12:14, 13 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 $$ (Error compiling LaTeX. )\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*} from which the answer is
~jeremylu (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
If you multiply both sides by
then it should come out to * = *
that then becomes * = *
which simplifies to
so now = putting in exponent form gets
=
so =
dividing yields and
- Pikachu13307
Solution 3
We can convert both and into and , respectively, giving:
Converting the bases of the right side, we get
Dividing both sides by , we get
Which simplifies to
Log expansion allows us to see that
, which then simplifies to
Thus,
And
-lepetitmoulin
Solution 4
is the same as
Using Reciprocal law, we get
~OlutosinNGA
Solution 5
. 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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