2022 AMC 12A Problem 14 — Solution | LIVE by Po-Shen Loh

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2022 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2022 AMC 12A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:logarithm sum and difference of cubes

Difficulty rating: 1730

14.

What is the value of

$(\log 5)^3+(\log 20)^3+(\log 8)(\log 0.25)$

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

$\dfrac32$

$\dfrac74$

$2$

$\dfrac94$

$3$

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 $a^3+b^3=(a+b)\big((a+b)^2-3ab\big)=2\big(4-3(1-u^2)\big)=2+6u^2.$

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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