Problem: When simplified logβ‘8Γ·logβ‘18\log 8 \div \log \dfrac{1}{8}log8Γ·log81β becomes:
Answer Choices:
A. 6logβ‘26 \log 26log2
B. logβ‘2\log 2log2
C. 111
D. 000
E. β1-1β1 Solution:
logβ‘8Γ·logβ‘18=logβ‘8Γ·(logβ‘1βlogβ‘8)\log 8 \div \log \dfrac{1}{8} =\log 8 \div(\log 1-\log 8)log8Γ·log81β=log8Γ·(log1βlog8) =logβ‘8Γ·(βlogβ‘8)=β1=\log 8 \div(-\log 8)=-1=log8Γ·(βlog8)=β1