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

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==Solution 3==
 
==Solution 3==
  
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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
+
We can estimate the solution. Using <math>\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9</math> and <math>\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,</math> we have
  
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<cmath>0.7^3 + 1.7^3 + ).9\cdot(-0.6) = \boxed{2}</cmath>
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<cmath>0.7^3 + 1.7^3 + .9\cdot(-0.6) = \boxed{2}</cmath>
 
~kxiang
 
~kxiang
 +
 
==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 21:37, 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 $\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9$ and $\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,$ we have

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

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

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