2018 AMC 12A Problems/Problem 14
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. ! Missing $ inserted.)\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 |
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