2
$\begingroup$

Following are the Equations: \begin{align} \log x +\frac {\log(xy^8)}{((\log x)^2+(\log y)^2)} &= 2 \\ \log y + \frac{\log(x^8/y)}{((\log x)^2 +(\log y)^2)} &= 0 \end{align}

I tried substituting $\log x$ and $\log y$ with $a$ and $b$ but it results in a cubic equation with two variables .

$\endgroup$
1
$\begingroup$

These are

$\log x +\dfrac{\log(x/y^8)}{(\log x)^2+(\log y)^2} = 2 $ and $\log y+\dfrac{\log((x^8)y)}{(\log x)^2 +(\log y)^2)} =0 $.

Expanding the logs,

$\log x +\dfrac{\log x-8\log y}{(\log x)^2+(\log y)^2} = 2 $ and $\log y+\dfrac{8\log x+\log y}{(\log x)^2 +(\log y)^2)} =0 $.

Letting $\log x = a, \log y = b$,

$a +\dfrac{a-8b}{a^2+b^2} = 2 $ and $b+\dfrac{8a+b}{a^2 +b^2)} =0 $.

The 8 is a mysterious constant, so replace it by $c$.

$a +\dfrac{a-cb}{a^2+b^2} = 2 $ and $b+\dfrac{ca+b}{a^2 +b^2} =0 $.

Now, solve.

Looks like we will get a cubic, like you wrote.

$2(a^2+b^2) =a(a^2+b^2) +a-cb $ and $0= b(a^2+b^2)+ca+b $.

From the first, $a^2+b^2 =\dfrac{cb-a}{a-2} $.

Putting this in the second,

$\begin{array}\\ 0 &=b\dfrac{cb-a}{a-2}+ca+b\\ &=\dfrac{b(cb-a)+(ca+b)(a-2)}{a-2}\\ &=\dfrac{cb^2-ab+ca^2-2ca+ab-2b}{a-2}\\ &=\dfrac{cb^2+ca^2-2ca-2b}{a-2}\\ \end{array} $

so, if $a \ne 2$, $0 =cb^2+ca^2-2ca-2b $.

If $a=2$, $0 =b(a^2+b^2)+ca+b =b(4+b^2)+2c+b $ so $0 =b^3+5b+2c $. For $c=8$, this has a negative real (about -1.8771) and two complex roots.

This looks like a mess, so I am probably doing something wrong, so I'll stop here.

$\endgroup$
  • $\begingroup$ Thanks for your time. Your simplifications are quite helpful. I will work it out more clearly. $\endgroup$ – Onkar Dahale Sep 15 '18 at 17:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.