2022 AMC 12A Problem 14

Attempt Problem 14 of the 2022 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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14.

What is the value of

$\begin{aligned} &(\log 5)^3+(\log 20)^3 \\ &\quad {}+(\log 8)(\log 0.25) \end{aligned}$

where $\log$ denotes the base-ten logarithm?

$\dfrac32$

$\dfrac74$

$2$

$\dfrac94$

$3$

Answer: C

Concepts:logarithm sum and difference of cubes

Difficulty rating: 1730

Solution:

Let $u=\log 2.$ Then $\log 5=1-u,$ $\log 20=1+u,$ $\log 8=3u,$ and $\log 0.25=-2u.$

With $a=1-u,\ b=1+u,$ we have $a+b=2$ and $ab=1-u^2,$ so $\begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2. \end{gathered}$

The last term is $(3u)(-2u)=-6u^2,$ so the total is $2+6u^2-6u^2=2.$

Thus, the correct answer is C.

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