Question

Please help me simplify this. I keep getting the wrong answer.

Answer 1

\[\frac{ 1 }{ 4 } \ln y - \frac{ 1 }{ 4 } \ln (4-y) = x + \frac{ 1 }{ 4 } \ln 3\]

Answer 2

I have to express y in terms of x.

Answer 3

\[\frac{ 1 }{ 4 } \ln \frac{ y }{ 4-y } = x + \frac{ 1 }{ 4 }\ln 3 \]
\[\frac{ 1 }{ 4 }\ln \frac{ y }{ 3(4-y) } = x\]
\[\frac{ y }{ 12-3y } = e ^{4x}\]

Answer 4

\[y = \frac{ 12e ^{4x} }{ 1+ 3e ^{4x} }\]

Answer 5

But that is not the answer. :'(
I'll post the original question then.
It gives \[\frac{ 1 }{ 4 } \ln y - \frac{ 1 }{ 4 } \ln (4-y)= x + c\]
and says that when x=0 and y=1, express y in terms of x.

Answer 6

when x=0 , y should b 2 ... not 1

Answer 7

No, the question gives these values so that we can find c.

Answer 8

kz .. i'll try it once more ..

Answer 9

\[\frac{ 1 }{ 4 } \ln \frac{ 1 }{ 3 } = c\]

Answer 10

Okay.

Answer 11

http://www.wolframalpha.com/input/?i=solve+y%2C+%5Cfrac%7B+1+%7D%7B+4+%7D+%5Cln+y+-+%5Cfrac%7B+1+%7D%7B+4+%7D+%5Cln+%284-y%29%3D+x+%2B+%5Cfrac%7B+1+%7D%7B+4+%7D+%5Cln+%5Cfrac%7B+1+%7D%7B+3+%7D+

Answer 12

still stuck ?

Answer 13

If the original problem says
\[\frac{ 1 }{ 4 } \ln y - \frac{ 1 }{ 4 } \ln (4-y)= x + c\]and asks us to find the value of \(c\) when \(x = 0\) and \(y = 1\):

\[\frac{1}{4}\ln 1 - \frac{1}{4}\ln(4-1) = 0+c\]\[0-\frac{1}{4}\ln 3 = c\]

Answer 14

I think the question is about finding the specific solution curve that passes through (0, 1)

Answer 15

\[\frac{ 1 }{ 4 } \ln y - \frac{ 1 }{ 4 } \ln (4-y)= x + c \]

plugin \(x=0,~~y=1\)
you get : \(c = \frac{ 1 }{ 4 } \ln \frac{ 1 }{ 3 } \)


\[\implies \frac{ 1 }{ 4 } \ln y - \frac{ 1 }{ 4 } \ln (4-y)= x + \frac{ 1 }{ 4 } \ln \frac{ 1 }{ 3 } \]
\[ \ln y - \ln (4-y)= 4x + \ln \frac{ 1 }{ 3 } \]
\[ \ln \dfrac{3y}{4-y} = 4x \]
\[ \dfrac{3y}{4-y} = e^{4x} \]
\[ y = \dfrac{4e^{4x}}{e^{4x}+3} \]

Answer 16

@rational :
The answer given is:
\[y=\frac{ 4 }{ 3e ^{-4x} +1 }\]

Answer 17

hahah both are equivalent

Answer 18

divide top and bottom by \(e^{4x}\)

Answer 19

\[y = \dfrac{4e^{4x}}{e^{4x}+3} = y = \dfrac{\dfrac{4e^{4x}}{e^{4x}}}{\dfrac{e^{4x}+3}{e^{4x}}}\]

Answer 20

simplify

Answer 21

>_<

Answer 22

Oh Lord. I have been trying since ever. I was like where did the 1 come from? I'm so sorry.

Answer 23

lol happens sometimes when u do nightouts in a row, no problem at all :)

Answer 24

I'll do the numerator and denominator separately:
Numerator:
\[\frac{4e^{4x}}{e^{4x}} = 4*\frac{e^{4x}}{e^{4x}} = 4*\frac{\cancel{e^{4x}}}{\cancel{e^{4x}}}=4*1=4\]
Denominator:
\[\frac{e^{4x}+3}{e^{4x}} = \frac{e^{4x}}{e^{4x}} + \frac{3}{e^{4x}} = 1 + \frac{3}{e^{4x}}=1+3e^{-4x}\]because \[\frac{1}{x^n} = x^{-n}\]

Numerator/denominator is therefore
\[\frac{4}{1+3e^{-4x}}\]