2007-09-22 14:29:59 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x=5+2ln(y)
x-5=2ln(y)
lny=(x-5)/2
y=e^(x-5)/2
2007-09-22 14:34:12 · answer #1 · answered by chasrmck 6 · 1⤊ 0⤋
switch the y and the x:
x= 5 +2ln(y)
isolate y:
(x-5)/2 =lny
raise both sides to the e:
e^ [(x-5)/2] =y
this is the inverse
2007-09-22 21:35:51 · answer #2 · answered by sayamiam 6 · 0⤊ 0⤋
y = 5 + 2ln x
y - 5 = ln x
1/2(y - 5) = ln x
x = exp[1/2(y - 5)]
2007-09-22 21:35:48 · answer #3 · answered by selang er uu 2 · 0⤊ 1⤋
y = 5 + 2ln(x)
Inverse
x = 5 + 2ln(y)
ln(y) = (x-5)/2
e^(ln(y)) = e^((x-5)/2)
y = e^((x-5)/2) = â(e^(x - 5))
2007-09-22 21:36:24 · answer #4 · answered by Marvin 4 · 0⤊ 0⤋
y = 5 + 2*ln(x)
y - 5 = 2 ln(x)
(y - 5)/2 = ln(x)
if a = b, then e^a = e^b
also, e^ln(x) = x
e^[(y-5)/2] = x
which you could leave like that (if you are programming a computer to do the operation for you) or:
x = e^[(y-5)/2]
x = SQRT[e^(y-5)]
x = SQRT[e^y / e^5]
2007-09-22 21:37:04 · answer #5 · answered by Raymond 7 · 0⤊ 0⤋
y = 5 + 2 ln(x)
y - 5 = 2 ln(x)
(y -5 ) / 2 = ln(x)
exp[ ( y - 5) / 2) ] = x
You could have done all that if you tried.
2007-09-22 21:33:55 · answer #6 · answered by morningfoxnorth 6 · 0⤊ 1⤋
y = 5 + 2 ln(x)
y-5=2 ln (x)
ln (x) = (y - 5)/2
x = e^((y-5)/2)
2007-09-22 21:38:38 · answer #7 · answered by ecosierra51 2 · 0⤊ 0⤋