Solve the system:?

Question

x^x+y = y^12
y^x+y = x^3

It is a system of equations.

Answer 1

There are at least TWO solutions [ ### Later : there's A THIRD SOLUTION ; see POSTSCRIPT 2 for information on THAT solution] :

Solution 1 : x = 4, y = 2, or

Solution 2 : x = 9, y = - 3.

[ ### PLUS Solution 3 : x ~ 0.027524663... , y ~ - 6.027524663... . ]

Here's how I got them:

The original equations SHOULD have been written *** [ See POSTSCRIPT 1 ] as

x^(x + y) = y^12 ......(A) and

y^(x + y) = x^3 ......(B).

Re-write these equations logarithmically, then:

(x + y) log x = 12 log y ......(C), and (x + y) log y = 3 log x ......(D).

[This isn't really ESSENTIAL, but it's more convenient to deal with everything "on the line" rather than "up in the exponential heavens."]

So, from (C) and (D),

(x + y) log x = [12 / (x + y)] (x + y) log y = [12 / (x + y)] 3 log x,

or (x + y)^2 = 36. ......(E)

Then x + y = 6 ......(F).

[ ### Later : The THIRD SOLUTION corresponds to the possibility x + y = - 6, with y negative. See POSTSCRIPT 2. ]

Check: then x^6 = y^12 ......(A') and y^6 = x^3 ......(B'), the revised and mutually consistent versions of eqns (A), (B).

These equations thus give x = y^2. (Equation (B') does not permit x = - y^2, which (A) would.) But since, from eqn (F), x + y = 6, we now have:

y^2 + y - 6 = 0, i.e. (y + 3) (y - 2) = 0. ......(G)

Therefore we have

Solution 1 : y = 2 (and x = 4) ......(H),

OR

Solution 2 : y = - 3 (and x = 9). ......(I)

CHECK these separate putative solutions:

1. Does 4^6 = 2^12 and 2^6 = 4^3 ? --- YES; and

2. Does 9^6 = (- 3)^ 12 and (- 3)^6 = 9^3 ? --- also YES.

(As implied above, the negative value for one of these y's suggests that I should have kept the solutions in their original form, or alternatively have taken care to write log [mod y] etc.)

Live long and prosper.

*** POSTSCRIPT 1 : Please note that whenever a POWER (or EXPONENT) is more complex than a SINGLE figure or symbol (and sometimes even then!), it should be ENCLOSED IN PARENTHESES OR BRACKETS, etc., so as to AVOID both possible ambiguities and/or confusing the reader. Every attempt possible should be made to communicate clearly with your intended reader(s). That is particularly necessary when writing good mathematics.

Puggy was quite right in querying what you meant, because you did not follow good practice. Fortunately, I was able to use my Vulcan telepathic mind-melding technique on you, and thus to divine what you really meant. Not all humans are so endowed, however.

### POSTSCRIPT 2 : I realised later that just ONE MORE SOLUTION is possible, corresponding to the alternative (negative) square root solution one can take in equation (E). The resulting value for the sum (x + y) differs only in sign from that already displayed in equation (F), namely:

x = y = - 6. ......(F ')

With this solution, we obtain the rather different (x, y) connection:

x = + y^(-2). ......(J)

(The second of the original equations, now y^(-6) = x^3, rules out the possibility x = - y^(-2), for example.)

This type of solution is harder to deal with than the simple integer solutions satisfying equation (F). In fact, instead of the factorizable quadratic equation (G), one now gets an essentially unfactorizable CUBIC equation (if one multiplies the next equation up by y^2, for example). From equations (F ') and (J) one gets:

f(y) = 1/y^2 + y + 6 = 0. ......(K)

There is NO positive y-solution for this. For positive y, 1/y^2 + y is always positive. It has one stationary point, a MINIMUM at y = 2^(1/3), value 3/2^(2/3) --- just < 2, so that f(y)'s own minimum for positive y is just under 8. In other words, for positive y, f(y) can NEVER equal zero.

However, there's a negative solution, and a quick, crude sketch --- plus the fact that 1/(-6)^2 is so small --- shows that the solution must be a little less than y = - 6, where the terms y + 6 themselves give zero and the value of f(y) is only 1/36. In fact the solution must clearly be near - [6 1/36] (!) --- or - 6.02777... . With that as a starting point, it's quick and easy to home in on this third solution, which is:

Solution 3 : x ~ 0.027524663... , y = - 6.027524663... ,

This gives f(y) ~ - 1.585 x 10^(-10), as close to zero as the successive additions with my calculator will allow.

Note also that 1/y^2 = 1/(- 6.027524663...)^2 = 0.027524663..., so that the requirement that x = 1/y^2 is indeed met.

This is the ONLY solution of this type ( for x + y = - 6 rather than + 6), as f(y) has a positive derivative for all negative y.

[ Note that there's a slight snag with my logarithmic versions of the equations --- I should have kept them in the exponential form, where the minor difficulty with having a negative argument for the log function wouldn't arise. One of my original TWO solutions already had a perfectly valid negative y, as does this new THIRD solution. I should presumably have allowed for log v to be log (mod v) where v could have been x and/or y. I went back to the original equations and checked that method, to be sure. Everything could in fact have been worked out properly in terms of the exponents of x and y, without ever going to what I thought would be the "simpler" logarithmic approach. ]

Answer 2

x^(x + y) = y^12
y^(x + y) = x^3

This isn't an easy system of equations to solve.
What I'm going to do is convert these two equations to logarithmic form.

log[base x](y^12) = x + y
log[base y](x^3) = x + y

By the log property log[base b](a^c) = c log[base b](a),

12 log[base x](y) = x + y
3 log[base y](x) = x + y

Seeing as we now have two equations equal to x + y, let's equate them to each other.

12 log[base x](y) = 3log[base y](x)

Divide both sides by 3,

4log[base x](y) = log[base y](x)

Use the log property that log[base b](a) = 1 / log[base a](b)

4log[base x](y) = 1 / log[base x](y)

Now multiply both sides of the equation by log[base x](y), and then divide by 4.

{ log[base x](y) }^2 = 1/4
log[base x](y) = +/- 1/2

This gives us two equations to solve for:
(a) log[base x](y) = 1/2 and (b) log[base x](y) = -1/2

Solving individually:

(a) log[base x](y) = 1/2 means
x^(1/2) = y, or
x = y^2

(b) log[base x](y) = -1/2 means
x^(-1/2) = y
1/x^(1/2) = y, so
1/x = y^2
x = 1/y^2
x = y^(-2)

(a) Plugging x = y^2 into the second equation,

y^(y^2 + y) = [y^2]^3
y^(y^2 + y) = y^6

Now, we can equate the exponents.

y^2 + y = 6, implying
y^2 + y - 6 = 0, giving us the solutions y = {-3, 2}

(b) Plugging x = y^(-2) into the second equation,

y^( y^(-2) + y ) = [y^(-2)]^3
y^( y^(-2) + y ) = y^(-6)

Equating the exponents,

y^(-2) + y = y^(-6)

Multiplying both sides by y^6,

y^4 + y^5 = 1, so
y^5 + y^4 - 1 = 0

This has a solution between 0 and 1; let's call this solution r.

By the combination of (a) and (b),
y = {-3, 2, r}

Got stuck at this point.

Answer 3

a million. sparkling up for between the variables. it truly is least perplexing to apply the second one equation. So contained in the second one equation upload y to both area and also you get x=4+y 2. Then, change that throughout for the x in equation1 to get 3(4+y)+2y=12 3. Distribute the three to get 12+3y+2y=12 4. Subtract 12 from both area to get 3y+2y=0.5. upload y's jointly to get 5y=0 6. Divide with assistance from 5 and y=0 7. Plug y=0 into the second one equation and also you get x=4 So, you've been precise! x=4 and y=0