I am trying to find number of solutions $(x,y)$ for :
$$\begin{align} 3^x +4^y &=13 \\ \log_3(x)-\log_4(y)&=1\end{align} $$
At first I tried :
$$\log_3(x)=1+\log_4(y)$$
and then :
$x=3^{\log_4(4y)}$
in the end : $3^{3^{\log_4(4y)}} +4^y =13$.
I am trying to find if this has a unique solution but i am stuck here.
Any help will be appreciated.
On
COMMENT.-We have $$\log_3(x)-\log_4(y)=1\iff\log_{3}(x)-\frac{\log_3(y)}{\log_3(4)}=\log_3(3)$$ then by injectivity of logaritms one has, putting $a=\log_3(4)$, $$x= 3y^{\dfrac 1a}$$ so the resultant equation $$27^{y^{\frac 1a}}+4^y=13$$ it is a trascendental equation which leads to the approximate solution $\boxed{(x,y)=(2.14,0.65)}$
In order to find the number of solutions of a problem with these constraints we firstly try to write each constraint in the style of functions: A: $$ 3^x + 4^y = 13 \Rightarrow y = \log_4 (13-3^x) = f(x) $$ B: $$ \log_3 (x) - log_4 (y) = 1 \Rightarrow y = 4^{log_3(x) - 1} = g(x) $$
for any solution we have: $$ 4^{log_3(x) - 1} = \log_4 (13-3^x) \Rightarrow f(x) = g(x) $$ clearly the function $g(x)$ is strictly increasing with an infimum of 0. and the function $f(x)$ is clearly strictly decreasing with a supremum of $\log_4 (13) > 0$ since any strictly increasing or strictly decreasing function is one-to-one, for any value of y in the range of f(x) and g(x)there is exactly one solution to $y = f(\alpha)$ and $y = g(\beta)$. Now first suppose there is no solution. that means only one thing : $$\alpha \neq \beta$$ this means that $h(x) = g(x) - f(x)$ has no solution for $h(x) = 0$ but since $g(x)$ is strictly increasing and $-f(x)$ is strictly increasing, $h(x)$ is strictly increasing too. but we know that the minimum of h(x) is certainly less than zero(can you say why?). and since h(x) is also one-to-one that means there is exactly one solution for $h(x)=0$ therefore, the problem has exactly one solution.