My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?(Smirk)

y = e^3

y = e^3

e = 3SQRT(y)

e = y^1/3

e^2 = y^2/3

Thanks(Nod)

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- August 14th 2009, 01:10 AMbobbyboy1111Incorrect equation, please help me fix it.
My teacher said this is incorrect because I should be taking the natural log of both sides of the equation. How would I do that?(Smirk)

y = e^3

y = e^3

e = 3SQRT(y)

e = y^1/3

e^2 = y^2/3

Thanks(Nod) - August 14th 2009, 01:48 AMfuror celtica
what are you trying to find? if you are looking for y, you have it already. e^3 is a constant.

- August 14th 2009, 01:52 AMQM deFuturo

remember that when you have y = b ^ x, this can be written as log_b y = x.

In this case, b = e, so log_e (y) = 3. Log to base e is written ln, so this becomes ln (y) = 3.

But, it's not clear to me what problem you are really trying to solve. Are you trying to solve for y? If so, just compute e^3 on your calculator, and that's the answer. - August 14th 2009, 03:04 AMbobbyboy1111
- August 14th 2009, 05:03 AMfuror celtica
no. natural log of e^3 is 3 (obviously)

natural log of y is lny (obviously)

there is nothing more to it in this case.(Bow) - August 14th 2009, 05:12 AMbobbyboy1111
- August 14th 2009, 05:30 AMAmer
- August 14th 2009, 05:48 AMmr fantastic
Posted and discussed here: http://www.mathhelpforum.com/math-he...n-correct.html

Thread closed.