$$\left\{\begin{array}{l} \ln(x+5)+y^2=10\\ x^2+y^2=9 \end{array}\right.$$

I have to solve this system of equations.
Let us assume:

$ x=ty$

The system becomes: $$\left\{\begin{array}{l} \ln(ty+5)+y^2=10\\ t^2y^2+y^2=9 \end{array}\right.$$


$\dfrac{\ln(ty+5)}{t^2y^2} = \dfrac{10}{9}$

How can I go on?

  • $\begingroup$ I would try reducing it as $ x^{2} +y^{2}= ln(x+5) +y^{2} -1$ implies $x^{2}=ln(x+5)-1$ and then trying a graphical approach $\endgroup$ Oct 10, 2015 at 15:04

2 Answers 2


We have $y=\pm \sqrt{9-x^2}$, so it is enough to find the solutions for $x$. Now $\log(x+5)=10-y^2=x^2+1$, so we have to solve the equation $$ \log(x+5)=x^2+1. $$ Here we can find help at MSE (e.g., a similar case is here).

  • $\begingroup$ We seem to think alike and btw its $ln$ instead of $log$ $\endgroup$ Oct 10, 2015 at 15:07
  • $\begingroup$ Is there any easy generalised rule to solve the equation $xe^{x^{2}}+5e^{x^{2}}+e=0$ ? $\endgroup$ Oct 10, 2015 at 15:13
  • $\begingroup$ @SujithZis I am so used to $\log$ for the natural logarithm with base $e$, which is the standard convention in analytic number theory... $\endgroup$ Oct 10, 2015 at 15:36
  • $\begingroup$ oh i didn't know that deeply sorry . $\endgroup$ Oct 10, 2015 at 15:44

As Dietrich Burde answered and Sujith Zis commented, eliminating $y^2$ from the second equation reduces to finding the zero's of $$f(x)=\log (x+5)-x^2-1$$ The derivative $$f'(x)=\frac{1}{x+5}-2 x$$ cancels for $$x_{\pm}=\frac{1}{2} \left(\pm 3 \sqrt{3}-5\right)$$ but $x_-$ has to be discarded since smaller than $-5$.

$$f(x_+)=-14+\frac{15 \sqrt{3}}{2}+\log \left(\frac{1}{2} \left(5+3 \sqrt{3}\right)\right)\approx 0.619244$$ and the second derivative test shows that this is a maximum.

Now, by inspection $$f(-1)=\log (4)-2\approx -0.613706$$ $$f(0)=\log (5)-1\approx 0.609438$$ $$f(1)=\log (6)-2\approx -0.208241$$ So, there are two roots : the first one between $-1$ and $0$, the second one between $0$ and $1$.

Since the second derivative $$f''(x)=-\frac{1}{(x+5)^2}-2$$ is always negative, thse are the only two roots for $x$.

At this point, you need a numerical method such as Newton to find their numerical values.

I am sure that you can take from here.


You could have a very good estimation of the roots if you expand the function as a Taylor series at $x=0$ (this is justified since we notice that the roots are significantly smaller than $5$); this will give $$f(x)=(\log (5)-1)+\frac{x}{5}-\frac{51 x^2}{50}+O\left(x^3\right)$$ Just by curiosity, plot the two functions on the same graph. From the solutions of the quadratic, I guess that one Newton iteration would be sufficient.

  • $\begingroup$ Could you please elaborate fourth step the one where you wrote $f(x_+)$ . $\endgroup$ Oct 10, 2015 at 15:50

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