Yes.

>>7761421
Is that denominator: [eqn] \ln (2) \left ( 2^{2y_n} + x 2^{y_n} \right ) [/eqn] or [eqn] \ln \left ( 2 \left ( 2^{2y_n} + 2^{y_n} \right ) \right ) [/eqn]

>>7761483
Yes

>>7761483
[eqn] \ln (2) \left ( 2^{2y_n} + x 2^{y_n} \right ) [/eqn]

>>7761483
Where do you see ln?

>>7761421
You can cancel a 2^y_n

>>7761809
denominator

>>7761789
In that case factor out a 2^{y_n} in both the numerator and denominator and cancel.

>>7761820
>>7761842
Ahhh, thanks for un-sticking me. Love when the answer has been staring me in the face the whole time.

So it boils down to:
y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)}

That form looks vaguely familiar to me, but maybe that's as far as I can get

>>7762262
[eqn]y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)}[/eqn]

>>7762270
[eqn] y_{n+1}=y_n\ -\frac{2\left(2^{y_n}-x\right)}{\ln (2)\left(2^{y_n}+x\right)} [/eqn]
omg. where can i get a plugin for LaTeX so this isn't so painful?

>>7762262
>>7762270
>>7762275
\frac{} {} not \frac{}{} or else LaTeX just wont work on here

(from the first post)
[math]y_{n+1}=y_n - \frac{2 \left(2^{y_n}-x\right)} { \ln (2) \left(2^{y_n}+x \right)}[/math]