How many $(x,y)$ solutions does the system $\begin{cases}3^x+4^y=13 \\ \log_3x - \log_4y=1\end{cases}$ have?

Question

How many $(x,y)$ solutions does the system $\begin{cases}3^x+4^y=13 \\ \log_3x - \log_4y=1\end{cases}$ have?

As I tried to solve this problem, I noticed that there is a single pair $(x,y)$ for which $3^x+4^y =13$ as both $f(x)=3^x$ and $g(x)=4^x$ are strictly increasing functions. So, at most, there is one possible solution. By this logic, I identified $x=2$ and $y=1$, but the second condition so to speak isn't met. It is obvious this solution is wrong (because there is indeed 1 valid solution), and I'd be interested in knowing where the logic if faulty.

Answer 1

Your argument for there only being one solution pair for the first equation is wrong. Yes, both f(x) and g(y) are increasing, but imagine you are at one (x,y) solution of it. If you increase x, then you could "correct" y by decreasing it, and get to another solution.

Answer 2

I know this isn't a complete answer but I think that it's interesting. So here it is

The red line represents $3^x+4^y=13$

The blue line represents $\log_3x-\log_4y=1$

Answer 3

We have the equations $$3^x + 4^y = 13 \tag{I}\label{eqI}$$ $$\text{and}, \log_3(x) - \log_4(y) = 1 \tag{II}\label{eqII}$$

If we rearrange equation $\eqref{eqI}$, we can get $$y = \log_4 (13 - 3^x) \tag{III}\label{eqIII}$$ Similarly with $\eqref{eqII}$, we can write it as $$\log_4 (y) = \log_3 \left(\frac{x}{3}\right) \tag{IV}\label{eqIV}$$ Taking the derivative on both sides in $\eqref{eqIII}$, we get $$\frac{dy}{dx} = \frac{-3^x\cdot \log_e (3)}{(13-3^x)\cdot \log_e (4)}$$ Since $(13 - 3^x) \gt 0$ as it is inside the logarithm, hence the whole derivative is negative for all values of $x$ in the domain and this is a decreasing function.

Now, if we take the derivative on both sides in $\eqref{eqIV}$, we get $$\frac{dy}{dx} = 4^{\log_3(x/3)}\cdot \left(\frac{3}{x}\right)\cdot \log_3(4)$$ Here, the domain is $x \gt 0$ since $x$ is inside the logarithm and in the denominator. Hence the derivative is positive for all $x$ in the domain.

Now, the first function decreases in $\mathbb{R}$ while the second increases for $x \gt 0$. On solving $\eqref{eqIII}$ for $y = 0$, you will see that the first function has a positive $X$-intercept. This means the curves must intersect at one point and hence there is a single solution to the given system of equations.

Answer 4

Write the first equation as $$y = f(x) = \log_4(13-3^x)$$ Write the second equation as $$y = g(x) = 4^{\log_3(x/3)}$$ Define: $$h(x) := f(x) - g(x)$$ Notice that $h$ is a continuous and defined function in $(0,\log_3(13))$. Here, we will only talk about this range.
Calculating the derivative is messy but not impossible: $$h'(x) = -\frac{\log_4(3)\cdot3^x}{13-3^x} - \frac{\log_3(4)\cdot4^{\log_3(x/3)}}{x}$$which is always negative in the specified range, i.e, $\color{blue}{h'(0) < 0}$ for $x \in (0,\log_3(13))$.

Now, it is easy to see that $f(1) >1$ and $g(1)<1$. So, $\color{blue}{h(1)>0}$.
Also, $$\lim_{x \to \log_3(13)^-}f(x) = -\infty$$ while $g(\log_3(13))$ is positive and bounded. So, $\color{blue}{h(x) \to -\infty}$ as $x \to \log_3(13)$. Since $h(x)$ is continuous in this range, strictly decreasing (from the derivative) and its value goes from positive to negative, unique $\exists x_0,y_0 \in (0,\log_3(13))$ such that: $$h(x_0) = 0 \iff f(x_0) = g(x_0) = y_0$$ or in other words, $h(x)$ has exactly one zero in the range. Graphically, $\boxed{x_0 \approx2.142}$ and $\boxed{y_0 \approx0.654}$.

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As JonathanZ pointed out in the comments, there are more than one $(x,y)$ satisfying $y = f(x)$. The case $(2,1)$ is only an integer solution, but there are infinite real solutions out there! Also, you said that both $3^x$ and $4^y$ are increasing, but the exponent is not the same. I can increase one variable and decrease the other in such a way that the sum is constant. You can compare this to, say $x+y = 0$. Both $x$ and $y$ are increasing functions, but that does not mean there is only one $(x,y)$.

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Now that I overcame my laziness, I'll post more accurate values (computed using WA of course): $$\boxed{x_0 \approx 2.14242966587709}\text{ and }\boxed{y_0 \approx 0.65387933788227}$$

Answer 5

From the second equation $\log_3 x=\log_4(4y)=k\implies x=3^k\,,\,y=4^{k-1}.$ Hence the first equation becomes $$f(k)=3^{3^k}+4^{4^{k-1}}=13\tag1$$ Now, $f'(k)=\ln^23\,\,3^{3^k}3^k+\ln^24\,\,4^{4^{k-1}}4^{k-1}>0$ for all $k\in\Bbb R.$ So, $f(k)$ is strictly increasing function and $(1)$ has a unique solution which is obviously in $[0,1].$ In fact, WA computed $k\approx0.693548$ and the given system has the unique solution $(x,y)\approx(2.14243,0.65388).$

Answer 6

Since the problem of the number of solutions has already been answered by other users, make the problem more general using $a$ instead of $13$.

The second equation gives $$y=4^{\frac{\log (x)}{\log (3)}-1}$$

Plugging it in the first one, we look for the zero of function $$f(x)=3^x+4^{4^{\frac{\log (x)}{\log (3)}-1}}-a$$ which not the nicest possible.

Consider instead $$g(x)=\log\left(3^x+4^{4^{\frac{\log (x)}{\log (3)}-1}}\right)-\log(a)$$ which is quite close to linearity (this is very good for any root finding method).

Using Newton method with $x_0=1$ (this is far away from the solution), the first iterate is $$x_1=\frac{4 \left(3+\sqrt{2}\right) \log (3) \log \left(\frac{a}{3+\sqrt{2}}\right)+\sqrt{2} \log ^2(4)+12 \log ^2(3) } { 4 \sqrt{2} \log ^2(2)+12 \log ^2(3)}$$ For $a=13$, this would already give $x_1=2.21806$

Doing the same with $x_0=e$, $x_1=2.15035$