log rules: If I have (1/2)ln|y| = -ln|x| + c ....?

[Idealistic]

If I have (1/2)ln|y| = -ln|x| + c, is that the same as:

y = 2(1/x)e[sup:28udi8d7]c[/sup:28udi8d7] or y = -2x(e[sup:28udi8d7]c[/sup:28udi8d7])?

 

[stapel]

I get something different. How did you arrive at your answers?

Please be complete. Thank you!

Eliz.

 

[Idealistic]

Well Im really shaky with logs, but what I did was:

(1/2)ln|y| = -ln|x| + c

(1/2)ln|y| = ln|x[sup:13i33jy0]-1[/sup:13i33jy0]| + c

|y|/2 = e[sup:13i33jy0]ln|x-1|[/sup:13i33jy0]+ c[/sup]

|y|/2 = x[sup:13i33jy0]-1[/sup:13i33jy0]e[sup:13i33jy0]c[/sup:13i33jy0]

y = (2/x)e[sup:13i33jy0]c[/sup:13i33jy0]

.

 

[Deleted member 4993]

Well Im really shaky with logs, but what I did was:

(1/2)ln|y| = -ln|x| + c

ln(?y) + ln(x) = c

ln(x?y) = c

x?y = e[sup:wdh4u155]c[/sup:wdh4u155]

?y = e[sup:wdh4u155]c[/sup:wdh4u155]/x

y = e[sup:wdh4u155]2c[/sup:wdh4u155]/x[sup:wdh4u155]2[/sup:wdh4u155]

(1/2)ln|y| = ln|x[sup:wdh4u155]-1[/sup:wdh4u155]| + c

|y|/2 = e[sup:wdh4u155]ln|x-1|[/sup:wdh4u155]+ c[/sup]

|y|/2 = x[sup:wdh4u155]-1[/sup:wdh4u155]e[sup:wdh4u155]c[/sup:wdh4u155]

y = (2/x)e[sup:wdh4u155]c[/sup:wdh4u155]

.