• Aug 14th 2009, 12:10 AM
bobbyboy1111
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)
• Aug 14th 2009, 12:48 AM
furor celtica
what are you trying to find? if you are looking for y, you have it already. e^3 is a constant.
• Aug 14th 2009, 12:52 AM
QM deFuturo
Quote:

Originally Posted by bobbyboy1111
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)

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.
• Aug 14th 2009, 02:04 AM
bobbyboy1111
Quote:

Originally Posted by QM 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.

My teacher told me to solve by taking the natural log of both sides of the equation, that's it, finito.

So, would this be correct?:

y = e^3
log_e(y) = 3
ln(y) = 3

Can this be simplified any further?(Smirk)

Thanks.
• Aug 14th 2009, 04:03 AM
furor 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)
• Aug 14th 2009, 04:12 AM
bobbyboy1111
Quote:

Originally Posted by furor 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)

So would that be:

y = e^3

ln(y) = 3?(Happy)
• Aug 14th 2009, 04:30 AM
Amer
Quote:

Originally Posted by bobbyboy1111
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)

$\displaystyle y=e^3$

$\displaystyle \log _e y = \log _e (e^3)$ note that $\displaystyle log_e = ln$

$\displaystyle \ln y = 3 \ln (e)$

$\displaystyle \ln y = 3$
• Aug 14th 2009, 04:48 AM
mr fantastic
Quote:

Originally Posted by bobbyboy1111
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)

Posted and discussed here: http://www.mathhelpforum.com/math-he...n-correct.html