log sqroot 6+ log sqroot 2/3= ? base of log is 2.

Logarithmic problem.
in Pre-Algebra Answers by Level 1 User (580 points)

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1 Answer

log2(sqrt(6)) +log2(sqrt(2/3))=log2[sqrt(6)*sqrt(2/3)]

=log2[sqrt(6*2/3)]

=log2[sqrt(4)]

=log2(2)

=1
by

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