Difference between revisions of "2022 AMC 12A Problems/Problem 14"
Sugar rush (talk | contribs) (Choice E is equal to 3, not 5/2) |
Hithere22702 (talk | contribs) |
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<math>\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3</math> | <math>\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>\text{log } 2 = x</math>. The expression then becomes <cmath>(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.</cmath> | Let <math>\text{log } 2 = x</math>. The expression then becomes <cmath>(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.</cmath> | ||
-bluelinfish | -bluelinfish | ||
| + | |||
| + | ==Solution 2== | ||
| + | Using sum of cubes | ||
| + | <cmath>(\log 5)^{3}+(\log 20)^{3}</cmath> | ||
| + | <cmath>= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})</cmath> | ||
| + | <cmath>= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})</cmath> | ||
| + | Let x = <math>\log 5</math> and y = <math>\log 2</math>, so <math>x+y=1</math> | ||
| + | |||
| + | The entire expression becomes | ||
| + | <cmath>2(x^2-x(2y+x)+(2y+x)^2)-6y^2</cmath> | ||
| + | <cmath>=2(x^2+2xy+4y^2-3y^2)</cmath> | ||
| + | <cmath>=2(x+y)^2 = \boxed{2}</cmath> | ||
| + | |||
| + | ~Hithere22702 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:06, 14 November 2022
Contents
Problem
What is the value of ![\[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\]](http://latex.artofproblemsolving.com/a/f/1/af1508c28c6d3e8f63ecd5c43998464209b05c65.png)
where 
denotes the base-ten logarithm?
Solution 1
Let 
. The expression then becomes ![\[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\]](http://latex.artofproblemsolving.com/2/2/3/223a30976aa281499351492cd728483a4e1456b3.png)
-bluelinfish
Solution 2
Using sum of cubes
![\[(\log 5)^{3}+(\log 20)^{3}\]](http://latex.artofproblemsolving.com/0/8/a/08a4319f2159439f5ad5508c8c5de2aa8e8dba6c.png)
![\[= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})\]](http://latex.artofproblemsolving.com/2/0/7/2074325451a4f90d6d79749f3e1d47a8eaa84abe.png)
![\[= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})\]](http://latex.artofproblemsolving.com/7/d/e/7de8d5ee2cfb76914cd7099210040eb9b53dd8d4.png)
Let x = 
and y = 
, so 
The entire expression becomes
![\[2(x^2-x(2y+x)+(2y+x)^2)-6y^2\]](http://latex.artofproblemsolving.com/5/e/4/5e46b1000679ebea29bea0016de5c5f3f1c28b16.png)
![\[=2(x^2+2xy+4y^2-3y^2)\]](http://latex.artofproblemsolving.com/1/b/2/1b2479a5de4184f7d0a9028cb3a8c19ebd3073dd.png)
![\[=2(x+y)^2 = \boxed{2}\]](http://latex.artofproblemsolving.com/7/c/8/7c83b8d9cc53bb9158a7d62ec9a1eed6a878138e.png)
~Hithere22702
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. 
