# Simultaneous log equations

I'm going through logarithms at the moment, and I can't solve this simultaneous equation:

$$\log x - \log 2 = 2\log y$$ $$x - 5y + 2 = 0$$

I've tried substituting both $x$ and $y$ to no avail:

$$\log \left(\frac{5y - 2}{2}\right) = \log y^2$$

or:

$$\log \left(\frac{x}{2}\right) = \log \left(\frac{x+2}{5}\right)^2$$

But I can't get passed that. Can someone point out what direction I need to go in?

• $\log$ is injective. – user121880 Jul 5 '14 at 16:55
• @Zircht Can you define what you mean by "injective"? – hohner Jul 5 '14 at 16:56
• It means that $\log(a)=\log(b)$ implies $a=b$. – user121880 Jul 5 '14 at 16:57

From the first equation we get $\frac x2=y^2$ so with the second equation we get
$$2y^2-5y+2=0,\quad y>0$$ can you take it from here?
From $$\log \left(\frac{x}{2}\right) = \log \left(\frac{x+2}{5}\right)^2$$ you can raise to the power $10$ on both sides and get $$\frac{x}{2} = \left(\frac{x+2}{5}\right)^2.$$