How to solve $2^x +3^y =7 $ and $2x-3y=1$

Question

How to solve $2^x +3^y =7 $ and $2x-3y=1$

It is easy to see that, $x=2 $ and $y=1$ are solutions to the equations.

But how can I solve it?

Answer 1

Because $2x-3y=1$, we know that $x = \frac{1+3y}{2}$. Then $2^x+3^y = 7$ becomes $$2^{(1+3y)/2}+3^y=7$$ so it suffices to solve the above equation. Observe that the function $$f(y) = 2^{(1+3y)/2}+3^y$$ is increasing (exponential functions with base $>1$ are increasing, and the sum of two increasing functions is increasing). In particular, it is a one-to-one function, so there is a unique $y$ for which $$2^{(1+3y)/2}+3^y=7.$$ You've already found that $y=1$ is such a solution, so it is the only solution. (And $x$ is determined uniquely from $y$ as above.)

Answer 2

The structure of complex solutions to this system looks interesting. Here are some of the $y$ values:

Answer 3

For the fun of it, let me make the problem more general $$2^x +3^y =k \qquad \text{and} \qquad 2x-3y=1$$ As @Robert Israel commented, make $z=2^x$ and,as he wrote, the problem is no< to find the zero of function $$f(z)=z+3^{-\frac 13} z^\alpha-k\qquad \text{where} \qquad \alpha=\frac{2 \log (3)}{3 \log (2)}$$ $\alpha$ is very close to $1$ (this makes $f(x)$ to be very close to a straight line) and then, for any $k$, we can approximate the solution using $$z_0=\frac{3 k}{3+3^{2/3}}$$ Making a single iteration of Newton method, we have $$z_1=\frac{z_0 \left(3 k+3^{2/3} (\alpha -1) z_0^{\alpha }\right)}{3^{2/3} \alpha z_0^{\alpha }+3 z_0}$$

Some results for a few values of $k$

$$\left( \begin{array}{cccc} k & z_0 & z_1 & \text{solution} \\ 1 & 0.59054 & 0.59758 & 0.59757 \\ 2 & 1.18108 & 1.17662 & 1.17662 \\ 3 & 1.77162 & 1.74860 & 1.74860 \\ 4 & 2.36217 & 2.31598 & 2.31597 \\ 5 & 2.95271 & 2.87992 & 2.87990 \\ 6 & 3.54325 & 3.44113 & 3.44109 \\ 7 & 4.13379 & 4.00005 & 4.00000 \\ 8 & 4.72433 & 4.55703 & 4.55695 \\ 9 & 5.31487 & 5.11230 & 5.11220 \\ 10 & 5.90541 & 5.66605 & 5.66592 \\ 11 & 6.49596 & 6.21843 & 6.21828 \\ 12 & 7.08650 & 6.76957 & 6.76939 \\ 13 & 7.67704 & 7.31957 & 7.31936 \\ 14 & 8.26758 & 7.86852 & 7.86827 \end{array} \right)$$