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

thanks

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- September 23rd 2009, 06:36 AMPandaNomiumIs there a way to further reduce this?
ln(y)=(2/3)*ln(x)

thanks - September 23rd 2009, 06:40 AMstapel
I am guessing that the text in the subject line is your question...?

What do you mean by "reducing" the equation? (Wondering) - September 23rd 2009, 06:40 AMenjam
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) - September 23rd 2009, 06:51 AMPandaNomium
thanks enjam, that's just what i was confused about. i didn't know that e^2/3lnx = x^2/3

:) - September 23rd 2009, 06:58 AMstapel
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.

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) - September 23rd 2009, 07:02 AMPandaNomium
so what would be the simplified version of: e^(2/3lnx)

- September 23rd 2009, 08:53 AMmathaddict
- September 24th 2009, 05:06 AMpacman
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.