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Solve the following equation system over the real numbers $$\begin{cases} x(1-\log_{10}(5))=\log_{10}(11-3^y)\\ \log_{10}(35-4^x)=y\log_{10}(9) \\ \end{cases} $$

For the functions in the above relations to be well defined, I've determined that $0<x<\log_4(35)$ and $0<y<\log_3(11)$. I also know that each of the equations(taken separately by itself) has exactly one solution because of their monotonicity (in each case one is strictly increasing and the other is strictly decreasing, so they can't have more than one intersection.)

By substituting $(1-\log_{10}(5))$ with $\log_{10}(2)$ in the first equation the unique solution for it is found to be $x=\log_2(11-3^y).$

I've tried to proceed further but to no avail and come here looking for help; I think that there is no real solution to the system but haven't been able to prove it.

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  • $\begingroup$ Base of the logarithm $lg$? $\endgroup$ – Robert Z Feb 16 at 11:22
  • $\begingroup$ 10 is the base of lg of course $\endgroup$ – Luca Pana Feb 16 at 11:22
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The system can be written as $$\begin{cases} \log_{10}(2^x)=\log_{10}(11-3^y)\\ \log_{10}(35-(2^{x})^2)=\log_{10}((3^{y})^2) \end{cases}$$ Now if $u:=2^x<\sqrt{35}$ and $v:=3^y<11$, we can throw away the logarithm and solve with respect to $u$ and $v$: $$\begin{cases} u+v=11\\ u^2+v^2=35 \end{cases}$$ Can you take it from here?

P.S. Yes, as you already noted there are no real solutions otherwise $$70=2(u^2+v^2)=(u+v)^2+(u-v)^2\geq(u+v)^2=121.$$

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Hint: It is $$2^x=11-3^y$$ and $$35-2^{2x}=3^{2y}$$ Can you proceed?

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Hint

We have that

$$x(1-\log(5))=\log(11-3^y)\iff 2^x=11-3^y$$ and

$$\log(35-4^x)=y\log(9) \iff 35-4^x=9^y.$$

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