A Nice Olympiad Math Algebra Problem | A Nice Exponential Equation

[@davidseed2939]

2^(xx)=2x square both sides 2^(2xx) =4xx subs a=2xx 2^a=2a take log base 2 a =1+log_2(a) a=1 or 2 x=1/√2 or 1

[@SALogics]

Very nice! ❤

[@DJ.Nihad6174]

x =1 is not unique. you lost one solution. you lost one case. 2^(x² -1) =x 2^x² /2 =x 1 /2 =x×2^-x² squaring both sides. 1 /4 =x²×2^-2x² multiplying by -2 both sides. -1 /2 =-2x²×2^-2x² -1 /2 =-2x²×(e^ln2)^-2x² -1 /2 =-2x²×e^(-ln2×2x²) -ln2 /2 =-2x²ln2×e^(-ln2×2x²) ln(1 /2) /1 /2 =-2x²ln2×e^(-ln2×2x²). apply both sides lambert W function. ln(1 /2) =-2x²ln2 -ln2 =-2x²ln2 2x² =1, x =1 /√2. we reject negative solution. verify this solution. 2^(1 /√2)² -1 =1 /√2 2^(1 /2 -1) =1 /√2 2^(-1 /2) =1 /√2. this is second solution. x1 =1 /√2. x2 =1.

[@bagumnazmapervin1663]

Yes correct value of x is one by root two

[@davidseed2939]

i dont think you need Lambert function. starting from -1/2=-2x^2 . 2^(2x^2) 1/2=a.2^- a , where a=2x^2 rearranging 2^a=2a by inspection a=1 and a=2 are solutions. this corresponds to x=1/√2 and x=1

[@SALogics]

Very nice! ❤

[@bagumnazmapervin1663]

I think one by root two is also a correct value of x in this equation please justify it

[@SALogics]

Yes, you are right!❤