Math Olympiad | A Nice Exponential Problem | 90% Failed to solve !!

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[RyanLewis-Johnson-wqxs]

3333^(4x)+3333^(2x)=11112222 x=0.5=1/2

[Quest]

Factoring no 11112222= 2×3×1111×1667 hence eqn turns to( t+3334)(t-3333)= 0 where t = 3333^ 2x

Gives x= 1/2 real soln

[rodicabrudea]

Y=-3334 is possible in complex numbers

[davidbrisbane]

Let y = 3333^(2x), then
3333^(4x) + 3333^(2x) = 11112222 becomes
y² + y - 11112222 = 0

Now as 11112222 is pretty large,
then y² ≅ 11112222
So, y ≅ 3000+ or -3000+

Now it's not hard to factorise 11112222.
11112222 = 2¹*3¹*11¹*101¹

Now let's assume that y is a positive integer as we might get lucky with this assumption.

Looking at 11112222 = 2¹*3¹*11¹*101¹, then y = 3*11*101 = 3, 333 looks like it is the only possible positive integer solution for y.

Then we verify that y = 3, 333 is a solution.

3, 333² + 3, 333 =
3, 333(3, 334) = 11112222, so
y = 3, 3333 is indeed a solution.

Using synthetic division, we can divide y - 3333 into y² + y - 11112222 and we find the other solution is y = - 3334.

It's easy to find x using y = 3333^(2x).

First notice 3333^(2x) = -3, 334 has no real solution (but it does have complex solutions).

Next notice that
3333^(2x) = 3333 means that 2x = 1, so x = ½.