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lnx + .5lny - 5lnz

First, make the coefficients into exponents:

lnx + ln(y^.5) - ln(z^5)

ln[xy^.5] - ln(z^5)

ln[(xy^.5)/z^5]

There you go!

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lnx + .5lny - 5lnz

First, make the coefficients into exponents:

lnx + ln(y^.5) - ln(z^5)

ln[xy^.5] - ln(z^5)

ln[(xy^.5)/z^5]

There you go!

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There are 2 interpretations of your question:

First: e^[lnx + lny]

=e^[ln(xy)]

=xy

Second: lny + e^(lnx)

=lny + x

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I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)

Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2

Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * y

Finally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)

Sorry if everything is formatted really badly, this is my first post on answers.com.

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xy - xy = 0

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xy + xy = 2xy

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