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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?

Angelo Mark
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  • Why do you think an elementary solution to the generalized problem should be possible? Mixing exponentials and polynomials and looking for solutions screams transcendental equations to me which more often than not will have non-elementary solutions. – JMoravitz Jul 16 '20 at 17:02
  • It's "really" more like a polynomial but with an irrational exponent. That is, if $z = 2^x$, after substituting $y = (2x-1)/3$ you get $$ z + 3^{-1/3} z^\alpha = 7$$ where $\alpha = (2/3) \log_2(3)$. – Robert Israel Jul 16 '20 at 22:39

3 Answers3

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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.)

Hayden
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  • "You've already found that $y=1$ is such a solution"... and if the numbers were ever so slightly different? What then? What if for whatever reason the OP had not recognized $y=1$ was a solution by inspection? The general solution of $a^{p(y)}+b^{y}=c$ for $y$ given $a,b,c,p$ doesn't exist with elementary functions. I read the question as asking how to have found $y=1$ in the first place without inspection. – JMoravitz Jul 16 '20 at 17:09
  • what if there are no clear solutions at all ? – Angelo Mark Jul 16 '20 at 17:19
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    @AngeloMark In the general case, I don't think there is an elementary solution. See, e.g., this old question. Numerically, you can always determine whether there is a solution or not, and if there is a solution, you can approximate it arbitrarily closely. But an exact solution is not likely except in particular cases. – Hayden Jul 16 '20 at 17:32
  • @Jmoravitz If the explicit solution exist, then it should be expressed with Lambert $W$. (my opinition) – lone student Jul 16 '20 at 17:55
  • @lonestudent No, it has nothing to do with Lambert W. – Robert Israel Jul 16 '20 at 18:24
  • @RobertIsrael I wanted to say the worst (if there exist)... Then there is no explicit solution. Am I right? If there is an exact solution, what kind of a function can this be? The only method I can know in exponential transcendental solutions is Lambert – lone student Jul 16 '20 at 19:03
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    As far as I know there is no closed form for the solutions, whether using Lambert W or not. – Robert Israel Jul 16 '20 at 20:11
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The structure of complex solutions to this system looks interesting. Here are some of the $y$ values:

enter image description here

Robert Israel
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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 $k$

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

Claude Leibovici
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