Solving a non-linear system

anemone

MHB POTW Director
Staff member
Hi members of the forum,

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

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

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

Thanks!

Ackbach

Indicium Physicus
Staff member
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.

topsquark

Well-known member
MHB Math Helper
Hi members of the forum,

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

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

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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Jester

Well-known member
MHB Math Helper
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 Last edited:

anemone

MHB POTW Director
Staff member
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.

MarkFL

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. 