Difference between revisions of "2022 AMC 12A Problems/Problem 14"
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==Solution 4(log bash)== | ==Solution 4(log bash)== | ||
− | Using log properties, we combine the terms to make our expression equal to <math>\log {( (5^{\log^2{5}}) \cdot (20^{\log^2{20}}) \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. By exponent properties, we separate the part with base <math>20</math> to become <math>20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}}</math>. Then, we substitute this into the original expression to get <math>\log {( (5^{\log^2{5}}) \cdot 20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) } = \log {( (100^{\log^2{5}}) \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. Because <math>100^{\log^2{5}} = 25^{\log{5}}</math>, and <math>\log^2{20}-\log^2{5} = (\log{20}+\log{5})(\log{20}-\log{5}) = \log{100}\cdot\log{4} = 2\log{4}</math>, this expression is equal to <math> \log {( 25 ^ {\log{5}} \cdot 400^{\log{4}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. | + | Using log properties, we combine the terms to make our expression equal to <math>\log {( (5^{\log^2{5}}) \cdot (20^{\log^2{20}}) \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. |
+ | By exponent properties, we separate the part with base <math>20</math> to become <math>20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}}</math>. Then, we substitute this into the original expression to get <math>\log {( (5^{\log^2{5}}) \cdot 20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) } = \log {( (100^{\log^2{5}}) \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. | ||
+ | Because <math>100^{\log^2{5}} = 25^{\log{5}}</math>, and <math>\log^2{20}-\log^2{5} = (\log{20}+\log{5})(\log{20}-\log{5}) = \log{100}\cdot\log{4} = 2\log{4}</math>, this expression is equal to <math> \log {( 25 ^ {\log{5}} \cdot 400^{\log{4}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }</math>. | ||
+ | We perform the step with the base combining on <math>25</math> and <math>400</math> to get <math>25 ^ {\log{5}} \cdot 400^{\log{4}} = 25 ^ {\log{5}-\log{4}} \cdot 10000^{\log{4}} = 25^{\log{\frac{5}{4}}}\cdot 256</math>. Putting this back into the whole equation gives <math>\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 8^{\log{\frac{1}{4}}})}</math>. | ||
+ | One last base merge remains - but <math>25\cdot 8</math> isn't a power of 10. We can rectify this by converting <math>8^{\log{\frac{1}{4}}}</math> to <math>(4^\frac{3}{2})^{\log{\frac{1}{4}}} = 4^{\log{ \frac{1}{8} }}</math>. | ||
+ | Finally, we complete this arduous process by performing the base merge on <math>\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 4^{\log{\frac{1}{8}}})}</math>. | ||
+ | We get <math>25^{\log{\frac{5}{4}}} \cdot 4^{\log{\frac{1}{8}}} = 25^{\log{\frac{5}{4}}-\log{\frac{1}{8}}} \cdot 100^{\log{\frac{1}{8}}} = 25^{\log{10}} \cdot \frac{1}{64} = \frac{25}{64}</math>. | ||
+ | Putting this back into that original equation one last time, we get <math>\log(256 \cdot \frac{25}{64}) = \log{100} = \boxed{2}</math>. | ||
~aop2014 | ~aop2014 | ||
Revision as of 14:46, 18 November 2022
Problem
What is the value of where denotes the base-ten logarithm?
Solution 1
Let . The expression then becomes
-bluelinfish
Solution 2
Using sum of cubes Let x = and y = , so
The entire expression becomes
~Hithere22702
Solution 3
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
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.