Difference between revisions of "2022 AMC 12A Problems/Problem 14"
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+ | ===Solution 1.1=== | ||
+ | (Elaboration & motivation behind Sol. 1) | ||
+ | |||
+ | Note that <math>5,20,8</math>, and <math>0.25</math> are all products and quotients of exponents of <math>2</math> and <math>5</math>; this strongly hints at some sort of major simplification. | ||
+ | |||
+ | Note that we can write all of the following expressions in the following ways: | ||
+ | |||
+ | \begin{align*} | ||
+ | \log5=\log\dfrac{10}2=\log10-\log2&=1-\log2\\ \log20=\log(2\cdot10)=\log2+\log10&=\log2+1\\ | ||
+ | \log8=\log\left(2^3\right)&=3\log2 \\ | ||
+ | \log0.25=\log\left(2^{-2}\right)&=-2\log2 | ||
+ | \end{align*} | ||
+ | |||
+ | Thus, let <math>a=\log2</math>, and proceed as in solution 1. | ||
+ | |||
+ | ~Technodoggo | ||
==Solution 2== | ==Solution 2== |
Revision as of 01:25, 25 September 2024
Contents
Problem
What is the value of where denotes the base-ten logarithm?
Solution 1
Let . The expression then becomes
-bluelinfish
Solution 1.1
(Elaboration & motivation behind Sol. 1)
Note that , and are all products and quotients of exponents of and ; this strongly hints at some sort of major simplification.
Note that we can write all of the following expressions in the following ways:
\begin{align*} \log5=\log\dfrac{10}2=\log10-\log2&=1-\log2\\ \log20=\log(2\cdot10)=\log2+\log10&=\log2+1\\ \log8=\log\left(2^3\right)&=3\log2 \\ \log0.25=\log\left(2^{-2}\right)&=-2\log2 \end{align*}
Thus, let , and proceed as in solution 1.
~Technodoggo
Solution 2
Using sum of cubes Let x = and y = , so
The entire expression becomes
~kempwood
Solution 3 (Estimates)
We can estimate the solution. Using and we have
~kxiang
Solution 4(log bash)
Using log properties, we combine the terms to make our expression equal to . By exponent properties, we separate the part with base to become . Then, we substitute this into the original expression to get . Because , and , this expression is equal to . We perform the step with the base combining on and to get . Putting this back into the whole equation gives . One last base merge remains - but isn't a power of 10. We can rectify this by converting to . Finally, we complete this arduous process by performing the base merge on . We get . Putting this back into that original equation one last time, we get . ~aop2014
Video Solution (Speedy)
https://www.youtube.com/watch?v=pai2A9FXI9U
~Education, the Study of Everything
Video Solution (Simple)
https://youtu.be/7yAh4MtJ8a8?si=9vbP5erdxlCLlG82&t=2957
~Math-x
See Also
2022 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.