Log4 4(7/17) - log4 3(21/86) + log4 48

Answer 1

We can break this down using the properties of logarithms:

log4 4(7/17) = log4 (7/17) because loga a = 1 for any base a
log4 3(21/86) = log4 3 + log4 (21/86) because loga (b*c) = loga b + loga c
log4 48 = log4 (4*12) = log4 4 + log4 12 = 1 + log4 12 because loga (b*c) = loga b + loga c and loga a = 1

Now we can substitute and simplify:

log4 (7/17) - log4 3 - log4 (21/86) + 1 + log4 12
= log4 [(7/17)/(3*(21/86)*4))] + log4 12 because loga b - loga c = loga (b/c)
= log4 [7/(3*17*21*4/86)] + log4 12
= log4 [7/(3*17*21/43)] + log4 12
= log4 [7*43/(3*17*21)] + log4 12
= log4 (1/3) + log4 12^1/2 because 7*43 = 301 and 21*3*17 = 1071, which have a common factor of 3
= log4 (12^1/2 / 3) + 1/2
= log4 (2) + 1/2 because 12^1/2 = sqrt(12) = 2*sqrt(3)
= 1/2 + log4 2

Therefore, log4 4(7/17) - log4 3(21/86) + log4 48 simplifies to 1/2 + log4 2.

Answer 2

3

Answer 3

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