MHB Solving a Non-Linear System: Approaches and Techniques

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The discussion focuses on solving a non-linear system of equations involving exponential and logarithmic functions. A key insight is that the equation can be simplified by substituting \(2x - y = 1\), leading to a new equation that can be analyzed for solutions. The point \((0, -1)\) is identified as a solution to the system, but its uniqueness is questioned. A suggestion is made to use calculus to demonstrate that the first equation has only one solution by analyzing the function's behavior. The problem is acknowledged as an Olympiad-level challenge, prompting a discussion on the appropriateness of the forum for such inquiries.
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Hi members of the forum,

I am given to solve the following non-linear system:

Solve $$(1+4^{2x-y})(5^{1-2x+y})=1+2^{2x-y+1}$$ and $$y^3+4x+\ln(y^2+2x)+1=0$$

I'm interested to know how you would approach this problem because I don't see a way to do so.

Thanks!
 
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Well, $2x-y=1$ solves the LH equation, by inspection. You could rewrite this as $2x=y+1$. Plugging this into the other equation yields
$$y^3+2y+2+ \ln(y^2+y+1)+1=0,$$
or
$$y^3+2y+3+ \ln(y^2+y+1)=0.$$
WolframAlpha shows a solution of $y=-1$, which you can see solves the second equation. So the point $(0,-1)$ solves the system. It may not be unique.
 
anemone said:
Hi members of the forum,

I am given to solve the following non-linear system:

Solve $$(1+4^{2x-y})(5^{1-2x+y})=1+2^{2x-y+1}$$ and $$y^3+4x+\ln(y^2+2x)+1=0$$

I'm interested to know how you would approach this problem because I don't see a way to do so.

Thanks!
There may be a "fancy" method to showing there is only one solution, and I don't have one. :)

I do have a very suggestive graph however, which should give an idea about how to prove it. (I zoomed out to some really high values and that green function just keeps looking like it's a straight line.)

-Dan
 

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One way to show that

$\displaystyle (1+4^{2x-y})(5^{1-2x+y})=1+2^{2x-y+1}$

has the only solution $2x-y = 1$ is to let $u = 2x-y$ so the first equations becomes

$F(u)=\displaystyle 5^{1-u} + 5 \left(\frac{4}{5}\right)^u - 2^{u+1}-1$

Clearly, $F(1) = 0$ as shown previously. To show that $F(u) \ne 0$ for other values of $u$ is to show that $F' < 0$ for all $u$.

Side bar: Since this is in the Pre-Algebra Algebra section, calculus is probably not assumed :-)
 
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I want to thank all of you for helping me with this tough problem. It takes very little time to arrive at the result if we approach the problem by inspection, and then try to prove the first equation has only one solution using the calculus. I appreciate all of the help and thanks to MHB particularly for providing the platform for us to ask for guidance in every maths problems that we encounter.

P.S. This problem is actually an Olympiad maths problem and thus, I am sorry for posting this in this sub-forum but I don't know where else I should post this; sorry if I have posted it in an inappropriate sub-forum.
 
anemone said:
...
P.S. This problem is actually an Olympiad maths problem and thus, I am sorry for posting this in this sub-forum but I don't know where else I should post this; sorry if I have posted it in an inappropriate sub-forum.

Hello anemone,

Personally I feel you chose the sub-forum in which to post this problem appropriately. It is after all an algebra problem, and Jester was merely commenting that the calculus could be used as a tool to show the uniqueness of the solution, but he was unsure whether this was a technique you would want to consider given you posted here. I don't think he was implying you posted incorrectly.

I know you are careful about where and how you post, so you can rest assured the staff here does not in any way think you are careless about where you have posted a problem. (Happy)
 
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