Question

×= 3^100÷[3^40×3^58+( 3^110×3^15)÷3^27+(4^57÷4^56-1^4)×3^97]×2+24

Answer 1

x= 3^100÷[3^40×3^58+( 3^110×3^15)÷3^27+(4^57÷4^56-1^4)×3^97]×2+24

x= 3^100÷[3^98+3^125÷3^27+(4^1-1)×3^97]×2+24

x= 3^100÷[3^98+3^98+3×3^97]×2+24

x= 3^100÷[3^98+3^98+3^98]×2+24

x=3^100÷[3^98*(1+1+1)]×2+24

x=3^100÷(3^98*3)×2+24

x=3^100÷3^99×2+24

x=3×2+24

x=6+24

x=30