1. ## Transform Equation

I need help to transform this for y,I am struck after so far.the formula is
x= 3+ ln[ y ] Any help greatly appreciated
1 - y

2. Originally Posted by Jaffa
I need help to transform this for y,I am struck after so far.the formula is
x= 3+ ln[ y ] Any help greatly appreciated
1 - y
i am not sure what you wrote. and it is probably why your question hasn't been answered yet. please review your posts when you make them to make sure they are understandable

did you mean $\displaystyle x = 3 + \ln \bigg( \frac y{1 - y} \bigg)$ ??

3. ## Transform Equation

Yes Jhevon,x= 3+ ln[y/1-y],iswhat I was trying to write, sorry I am new to
posting on forums, but will try to get the hang of it. Hope you can help.
Thanks.

4. Originally Posted by Jaffa
Yes Jhevon,x= 3+ ln[y/1-y],iswhat I was trying to write, sorry I am new to
posting on forums, but will try to get the hang of it. Hope you can help.
Thanks.
ok, i will start by giving you this rule. it is a very important rule (the definition of a logarithm)

$\displaystyle \log_a b = c \Longleftrightarrow a^c = b$

note that $\displaystyle \ln x \equiv \log_e x$

can you apply that rule here to get rid of the log?

5. ## Transform Equation

Hi Jhevon
Iwill show you how far I have got,

x=3+ln[y/1-y]
x-3=ln[y/1-y
e^x-3=y/1-y
e^x-3(1-y)=y

Thanks

6. Originally Posted by Jaffa
Hi Jhevon
Iwill show you how far I have got,

x=3+ln[y/1-y]
x-3=ln[y/1-y
e^x-3=y/1-y
e^x-3(1-y)=y

Thanks
please use parentheses. you should have typed something like e^(x - 3) * (1 - y) = y

but yes, you are right so far. now expand the brackets

$\displaystyle e^{x - 3} - ye^{x - 3} = y$

you think you can take it from here? remember, to solve for $\displaystyle y$ we need to get all $\displaystyle y$'s on one side of the equation. then see if we can get to a $\displaystyle y = \cdots$ kind of format

7. Hi Jhevon
Still getting use to this posting,but learning !
I don't know if Iam right ,but here goes

e^(x-3)-ye^(x-3)=y
e^(x-3)-y=y/e^(x-3)
e^(x-3)=2y/e^(x-3)
e^(x-3)^2/2=y

Can't find away of writing inaformat like you do ?

8. Originally Posted by Jaffa
Hi Jhevon
Still getting use to this posting,but learning !
I don't know if Iam right ,but here goes

e^(x-3)-ye^(x-3)=y
e^(x-3)-y=y/e^(x-3)
e^(x-3)=2y/e^(x-3)
e^(x-3)^2/2=y
that's wrong. you cannot divide only one term on the right by $\displaystyle e^{x - 3}$.

as i said, get all the $\displaystyle y$'s together.

$\displaystyle \Rightarrow e^{x - 3} = y + e^{x - 3}y$

any ideas now?

Can't find away of writing inaformat like you do ?

see post #1 here

9. ## Transform Equation

Hi Jhevon
I am getting there ,I think

e^(x-3)=y+e^(x-3)y
e^(x-3)/e(x-3)=2y
then is ite^(x-3)-e^(x-3)/2=y

Maybe ?

10. Originally Posted by Jaffa
Hi Jhevon
I am getting there ,I think

e^(x-3)=y+e^(x-3)y
e^(x-3)/e(x-3)=2y
then is ite^(x-3)-e^(x-3)/2=y

Maybe ?
no

$\displaystyle e^{x - 3} = y + e^{x - 3}y$ ....................factor out $\displaystyle y$

$\displaystyle \Rightarrow e^{x - 3} = y(1 + e^{x - 3})$ ............divide by $\displaystyle 1 + e^x$

$\displaystyle \Rightarrow \frac {e^{x - 3}}{1 + e^{x - 3}} = y$