AMC 12A 2022 (Problem 14)FunctionsHardWhat is the value of (log5)3+(log20)3+(log8)(log0.25)(\log 5)^3+(\log 20)^3+(\log 8)(\log 0.25)(log5)3+(log20)3+(log8)(log0.25), where log\loglog denotes the base-ten logarithm?(A) 32\text{(A)}\;\frac32(A)23(B) 74\text{(B)}\;\frac74(B)47(C) 2\text{(C)}\;2(C)2(D) 94\text{(D)}\;\frac94(D)49(E) 3\text{(E)}\;3(E)3Related TopicsCoreLogarithmsMajorAlgebraic IdentitiesCheck AnswerYour answer:ABCDECheckHints (7)Hint 1Let x=log2x=\log 2x=log2. Find log5\log 5log5, log20\log 20log20, log8\log 8log8, and log0.25\log 0.25log0.25 in terms of xxx.Hint 2log20=log10+log2=1+log2=1+x\log 20=\log 10+\log 2=1+\log 2=1+xlog20=log10+log2=1+log2=1+xHint 3log8=log23=3log2=3x\log 8=\log 2^3=3\log 2=3xlog8=log23=3log2=3xHint 4log0.25=log14=log2−2=−2log2=−2x\log 0.25=\log\frac14=\log 2^{-2}=-2\log 2=-2xlog0.25=log41=log2−2=−2log2=−2xHint 5log5=log102=log10−log2=1−log2=1−x\log 5=\log\frac{10}{2}=\log 10-\log 2=1-\log 2=1-xlog5=log210=log10−log2=1−log2=1−xHint 6(1−x)3+(1+x)3+(3x)(−2x)=?(1-x)^3+(1+x)^3+(3x)(-2x)=?(1−x)3+(1+x)3+(3x)(−2x)=?Hint 7(1−x)3+(1+x)3+(3x)(−2x)(1-x)^3+(1+x)^3+(3x)(-2x)(1−x)3+(1+x)3+(3x)(−2x)=(1−3x+3x2−x3)+(1+3x+3x2+x3)−6x2=2=(1-3x+3x^2-x^3)+(1+3x+3x^2+x^3)-6x^2=2=(1−3x+3x2−x3)+(1+3x+3x2+x3)−6x2=2Final Answer(C) 222Related Problems (10)AMC 12B 2025 (Problem 7)AMC 12B 2024 (Problem 8)AMC 12A 2024 (Problem 8)AMC 12A 2023 (Problem 19)AMC 12A 2025 (Problem 7)AIME I 2026 (Problem 6)AMC 12B 2024 (Problem 15)AIME II 2025 (Problem 4)AMC 12B 2025 (Problem 24)AMC 12B 2025 (Problem 8)View all related problems →