# Solution for Simultaneous Logarithmic Equations

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 .

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.

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