A tricky problem from Harvard University Interview | Find X..?

A tricky problem from Harvard University Interview | Find X..? #maths #exponents #math

[@marklyles3335]

Poetry in Logs!!

[@ganeshdas3174]

2*4^x =16 =2*8
4^x = 8
2^2x = 2^3
2x = 3
x = 3/2

[@walterwen2975]

2(4^x) = 16, 4^x = 8, 2^2x = 2^3, 2x = 3; x = 3/2

[@RealQinnMalloryu4]

{4x+4x ➖ }+{4x+4x ➖ }={8x^2+8x^2}=16x^4 8^8x^4 4^4^4^4x^4 2^2^2^2^2^2^2^2x&2^2 1^1^1^1^1^1^1^1x^1^2 x^1^2 (x ➖ 2x+1).