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:

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


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

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

  • 4
    $\begingroup$ $\log$ is injective. $\endgroup$ – Zircht Jul 5 '14 at 16:55
  • $\begingroup$ @Zircht Can you define what you mean by "injective"? $\endgroup$ – hohner Jul 5 '14 at 16:56
  • 1
    $\begingroup$ It means that $\log(a)=\log(b)$ implies $a=b$. $\endgroup$ – Zircht 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?

  • $\begingroup$ Thanks - how obvious! I feel very silly now :) $\endgroup$ – hohner Jul 5 '14 at 16:59
  • $\begingroup$ You're welcome. Never mind it's good to try! $\endgroup$ – user63181 Jul 5 '14 at 17:00

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


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.