2022 AMC 12A Problems/Problem 14

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

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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2022 AMC 12A Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See Also