A Nice Algebra Problem | Math Olympiad | Can you solve this?

[@davidseed2939]

Looking at the equation the rhs is likely >1 so x is negative Subs x=-a 2^a= 6-a Now lhs is monotonic Increasing. and rhs is monotonic decreasing so there is one solution 2^a= 6-a a=2 is a solution x=-2 is the only solution.

[@SALogics]

Very nice! ❤

[@ilyashick3178]

Guessing 2^-1*(x+6)^-1 =1 >>>> can be solution as well.

[@SALogics]

Very nice! ❤

[@ilyashick3178]

@@SALogics Learning from your lessons for using function of W in right direction. Thanks !

[@MananArhamOP]

Big fan

[@SALogics]

Thank you so much! ❤

[@zawatsky]

2^(-x)=x+6⇔2^(-x)-x=6=4+2=2²+2. -x:=t. 2^t+t=2²+2⇔t=2⇔x=-2.

[@SALogics]

Very nice! ❤

[@wes9627]

(1/2)^{-2}=4

[@SALogics]

Very nice! ❤

[@adolphusduo3960]

At the second step, why did you write 1 instead of 1 raised to the power x?

[@SALogics]

Any power of 1 = 1 ❤

[@pspprabhat]

2^-×=×+6 =>1=(×+6)e^ln2(×) =>2^6.=(×+6)Ln e^ln2(×).e^Ln2(6) =>W{4ln2.e^ln2^4} ={(×+6)Ln 2 .e^ln2(×+6)} From W(a.e^a)=a 4ln2=×+6)Ln2 =>4-6=×=-2

[@SALogics]

Very nice! ❤