a(sqrt(asqrt(a)))=2 a^2(asqrt(a))=4 a^6*a=16 a^7-16=0 Solution 1: a=2^(4/7) Solution 2: on the complex plane, n^7 has to be 16, or in other words, 7θ=2pi θ=2pi/7, r=2^(4/7); a=2^(4/7)(cos(2pi/7)+isin(2pi/7))
These extra complex solutions do not exist. The only solution is 2^(4/7). You created more solutions when you squared both sides twice. The other replier was correct to only consider a^(7/4)=2 => a=2^(4/7) If you try plugging in those complex values, you will not get 2, but rather values that result in 2^4 when raised to the 4th power, e.g. -2,2i, and -2i. You must be careful to consider these cases when raising both sides to a power greater than 1.
@@n00bxl71 i though so, but based on his description, i decided to accept his as the correct answer. I've ignored complex and negative solutions way too many times lol.
a(sqrt(asqrt(a)))=2
a^2(asqrt(a))=4
a^6*a=16
a^7-16=0
Solution 1: a=2^(4/7)
Solution 2: on the complex plane, n^7 has to be 16, or in other words, 7θ=2pi
θ=2pi/7, r=2^(4/7);
a=2^(4/7)(cos(2pi/7)+isin(2pi/7))
Can't believe they call this an Olympiad question.
Though i didn't consider the complex plane while solving in mind. I just did a^7/4=2 so a=2^4/7 and called it quits
These extra complex solutions do not exist. The only solution is 2^(4/7). You created more solutions when you squared both sides twice. The other replier was correct to only consider a^(7/4)=2 => a=2^(4/7)
If you try plugging in those complex values, you will not get 2, but rather values that result in 2^4 when raised to the 4th power, e.g. -2,2i, and -2i. You must be careful to consider these cases when raising both sides to a power greater than 1.
@@n00bxl71 i though so, but based on his description, i decided to accept his as the correct answer. I've ignored complex and negative solutions way too many times lol.