Respuesta :

Kamikazeminem

[tex] ln( \sqrt{ex {y}^{2} {z}^{5} } ) \\ = ln{(ex {y}^{2} {z}^{5} ) }^{ \frac{1}{2} } \\ = \frac{1}{2} ln(ex {y}^{2} {z}^{5} ) \\ = \frac{1}{2} ln(e) + \frac{1}{2} ln(x {y}^{2} {z}^{5} ) \\ = \frac{1}{2} + \frac{1}{2} ln(x {y}^{2} {z}^{5} ) [/tex]

jcherry99

Answer:

Answer A  B and D

Step-by-step explanation:

This one requires and eagle eye. You have to be a bit careful with it.

The one you checked (D) is correct. There are a couple more and one is really tricky.

The first one is actually correct.

So is the second one, which I'll show first. Only C is incorrect.

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B]

ln(e^(1/2) + ln(x^(1/2)) + ln(y^(2/2) + ln(z^(5/2))

1/2 + ln(x^(1/2)) + ln(y) + (5/2) ln(z)

1/2 + 1/2 ln(x) + ln(y) + 5/2 ln(z)

A]

(1/2) * Ln(ex*z) + ln(yz^2)

(1/2)ln(e) + (1/2)ln(x) + (1/2)ln(z) + ln(y) + lnz^2

ln(e) = 1;   ln(z^2) = 2 ln(z)

1/2 + 1/2 ln(x) + 1/2 ln(z) + ln(y) + 2ln(z)

1/2 ln(z) + 2ln(z) = 5/2 ln(z)

So put all this together

1/2 + 1/2 ln(x) + 5/2 ln(z) + ln(y) which is Exactly like B

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