How to solve simultaneous exponential equations with polynomial parts? I have been puzzled at how to simultaneously solve the following equations and before I give up entirely I thought I'd turn to fellow mathematicians first:
$$2^x=3y$$
$$2^y=5x$$
I have graphed both of these and there are two solutions for both variables but I feel I have exhausted all of my own methods to find solutions algebraically. My ability is of a first year Physics undergraduate. Perhaps I have missed a trick, or they don't have enough information on their own, but whatever it is I'd be very grateful for solutions with a method to solve.
 A: You can't solve that analytically.
Eliminating $y$,
$$5x=2^{2^x/3}.$$
From the plot, you see two real roots, one between $0$ and $0.5$ ($0.2639479321406\cdots$), the other between $3.5$ and $4$ ($3.6521440237168\cdots$).
As you can check,
$$5\cdot 0<2^{1/3},5\cdot0.5>2^{\sqrt2/3},\\
5\cdot3.5>2^{8\sqrt2/3},5\cdot4<2^{16/3}.$$
The derivative seems monotonic, confirming a single extremum hence two roots.

Taking the logarithm twice and rearranging,
$$\ln(2)x=\ln(\ln(x)+\ln(5))-\ln(\ln(2))+\ln(3).$$
The extremum is found by canceling the derivative
$${\ln(2)}=\frac{\dfrac1x}{\ln(x)+\ln(5)},$$ 
or 
$$5x\ln(5x)=\frac5{\ln(2)}.$$
The function $5x\ln(5x)$ is monotonous increasing for $5x>\frac1e$ and there is a single solution.
A: Equations that mix exponential and polynomial terms, can't in general be solved. There are a few special cases, but this doesn't appear like one of them.
We can manipulate the equations a bit, $$2^x = 3y$$ $$2^y = 5x$$ gives $x = 2^y/5$ and $2^{\frac{2^y}{5}} = 3y$.
Thankfully, $x$ is a 1-1 function of $y$, so once $y$ is determined, $x$ is determined simply by plugging in $y$.
To solve the last equation: $2^{\frac{2^y}{5}} = 3y$, there are a few methods you can use to derive an approximate solution. The most straight forward method is to graph $f(y) = 2^{\frac{2^y}{5}} - 3y$ and estimate it's zeros to obtain an estimate of $y$.
If you want something you can do by hand, it will take longer, you can try to implement Newton's method. Newton's method is an iterative method, so you can achieve more and more accurate estimates through repeated application.
First you make a guess at the solution, say $y=1$. Then you take the tangent line of $f$ at $y=1$ and follow that line to zero. This zero becomes your first estimate, and your next guess.
To wit,
$$f'(y) = \frac{2^y}{5} 2^{2^y/5} - 3$$
$$f'(1) = \frac{2}{5} \cdot 2^{2/5} - 3$$
$$f(1) = 2^{2/5} - 3$$
and $$L(y) = (\frac{2}{5} \cdot 2^{2/5} - 3)(y-1) + ( 2^{2/5} - 3).$$
Thus $y_2$ solves $$L(y_2) = 0 = (\frac{2}{5} \cdot 2^{2/5} - 3)(y_2-1) + ( 2^{2/5} - 3)$$ and $$y_2 = \frac{-( 2^{2/5} - 3)}{(\frac{2}{5} \cdot 2^{2/5} - 3)} + 1$$ is our first guess at a solution for $y$.

Since there are two zeros to this function (we can see this by looking at the graph) the choice of initialization of Newton's method (ours was $y=1$ here) will lead to convergence to a different zero (if it does converge).
Graphically, it appears that $y\approx 0.40$ and $y\approx 4.19$ are solutions.
