Newton's Nightmares

The question isn't tricky

All that it requires is a determination

By the condition,

$\log{(3\cdot C_{m}^3)}-\log{C_{m}^1}=1,$

or

$\log{\frac{3C_{m}^3}{C_{m}^1}}=\log{10};$

from here,

$\frac{3C_{m}^3}{C_{m}^1}=10.$

Simplifying, we obtain the expression $m^2-3m-18=0$ , whose roots we get to be $m_1=6$ and $m_2=-3$ . The root appropriate to the binomial is $m=6$ .

From the condition that $9T_3-T_5=240$ we obtain the equation

$9C_{6}^22^{2(\frac{2}{3}+\frac{x}{2})}2^{4(\frac{x-1}{2}-\frac{1}{3})}-C_{6}^42^{4(\frac{2}{3}+\frac{x}{2})}2^{2(\frac{x-1}{2}-\frac{1}{3})}=240,$

from which we find

$9\cdot 2^{3x-2}-2^{3x+1}=16,$

or

$\frac{9\cdot 2^{3x}}{2^2}-2^{3x}\cdot2=16.$

And so,

$2^{3x}=2^6$

thus,

$\boxed{x=2}$