# Is there a way to further reduce this?

• Sep 23rd 2009, 05:36 AM
PandaNomium
Is there a way to further reduce this?
ln(y)=(2/3)*ln(x)

thanks
• Sep 23rd 2009, 05:40 AM
stapel
I am guessing that the text in the subject line is your question...?

What do you mean by "reducing" the equation? (Wondering)
• Sep 23rd 2009, 05:40 AM
enjam
ln(y)=(2/3)*ln(x)
= ln(y) = ln(x)^(2/3)
Multiplying both sides by e, we end up with:
y = x^(2/3)
• Sep 23rd 2009, 05:51 AM
PandaNomium
thanks enjam, that's just what i was confused about. i didn't know that e^2/3lnx = x^2/3

:)
• Sep 23rd 2009, 05:58 AM
stapel
Quote:

Originally Posted by enjam
Multiplying both sides by e, we end up with:
y = x^(2/3)

Actually, no. If you multiply through by a constant, you will end up with the same logarithmic equation, but now multiplied through by a constant. (Blush)

Instead, you need to raise both sides of the equation as powers on the base e.

Quote:

Originally Posted by PandaNomium
i didn't know that e^2/3lnx = x^2/3

There's a good reason for not having know that: it's not true! (Wondering)

However, e^ln(x^(2/3)) does equal x^(2/3). (Wink)
• Sep 23rd 2009, 06:02 AM
PandaNomium
so what would be the simplified version of: e^(2/3lnx)
• Sep 23rd 2009, 07:53 AM
Quote:

Originally Posted by PandaNomium
so what would be the simplified version of: e^(2/3lnx)

HI

e^[(2/3)lnx] would be x^(2/3)
• Sep 24th 2009, 04:06 AM
pacman
property of logarithm: p(ln x) = ln (x)^p

ln(y)=(2/3)*ln(x) = ln (x)^(2/3)

raise BS to e,

e^( ln y) = e^(ln (x)^(2/3))

but by definition, e^(ln y) = y, then

y = x^(2/3)

if you want a graph, see below.