Difference between revisions of "2022 AMC 12A Problems/Problem 14"

Revision as of 21:35, 16 November 2022

Problem

What is the value of $(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)$ where $\log$ denotes the base-ten logarithm?

$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$

Solution 1

Let $\text{log } 2 = x$ . The expression then becomes $(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.$

-bluelinfish

Solution 2

Using sum of cubes $(\log 5)^{3}+(\log 20)^{3}$ $= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})$ $= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})$ Let x = $\log 5$ and y = $\log 2$ , so $x+y=1$

The entire expression becomes $2(x^2-x(2y+x)+(2y+x)^2)-6y^2$ $=2(x^2+2xy+4y^2-3y^2)$ $=2(x+y)^2 = \boxed{2}$

~Hithere22702

Solution 3

We can estimate the solution. Using $\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) \approx 1.3, \log(8) \approx 0.9$ and $\log(.25) = \log(1)-\log(4) \approx -0.6,$ we have

$0.7^3 + 1.7^3 + ).9\cdot(-0.6) = \boxed{2}$ ~kxiang

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 23: / Line 23: @@
 ~Hithere22702
+==Solution 3==
+We can estimate the solution. Using <math>\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) \approx 1.3, \log(8) \approx 0.9</math> and <math>\log(.25) = \log(1)-\log(4) \approx -0.6,</math> we have
+<cmath>0.7^3 + 1.7^3 + ).9\cdot(-0.6) = \boxed{2}</cmath>
+~kxiang
 ==See Also==
 {{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}}
 {{MAA Notice}}

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