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# The logarithmic expression 3logbase2x - logbase2(2y) + 5 can be written?

a.) logbase2(3x^3/2y)

b.) logbase2(x^3/4y)

c.) logbase2 (16x^3/y)

d.) logbase2 (x^3/y^2) +5

### 4 Answers

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- la consoleLv 76 months ago
= 3.Log[2](x) - Log[2](2y) + 5 → recall: Log[a](x) = Ln(x) / Ln(a) ← where a is the base

= [3.Ln(x) / Ln(2)] - [Ln(2y) / Ln(2)] + 5

= { [3.Ln(x) - Ln(2y)] / Ln(2) } + 5 → recall: a.Ln(x) = Ln(x^a)

= { [Ln(x³) - Ln(2y)] / Ln(2) } + 5 → recall: Ln(a) - Ln(b) = Ln(a/b)

= { [Ln(x³/2y)] / Ln(2) } + 5 → recall: Log[a](x) = Ln(x) / Ln(a) ← where a is the base

= Log[2](x³/2y) + 5

- VamanLv 76 months ago
None seem to be the correct answer. If the question is 3 log base 2 x- 2 log base 2 y+5 then d is the correct answer.

- L. E. GantLv 76 months ago
Using log for log (base 2)....

3 log(x) - logf(2y) + 5

= log(x^3) - log(2y) + log(32)

= log(32x^3/(2y))

= log(16x^3/y)

- Anonymous6 months ago
Schools over U dumko

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