wildcard ssl certificate
2023 Mathematics Competition

Question 14

  • Mathematics
  • Maths
  • Competition
  • 2023

10 comments

Caleb Adeleye
1 year ago

@Akinola please check your wallet

Congratulations!
@Akinola Ajayi, you got it

Rose James
1 year ago

2²^y +2^y+1-8=0
(2^y)²-2^y ×2¹-8=0
Let p =2^y
(P)²-(p×2) -8=0
P²-2p-8=0
P²-4p+2p-8=0
P(p-4)+2(p-4)=0
(P+2) (P-4) =0
P+2=0 or p-4=0
P=-2 or p=4
2^y =2²
Y=2

Victory Light
1 year ago

2^2y-8 = -2^y-1 clt
2^2y + 2^y-1 = 8
Apply law of indices remove same base
2y + y-1 = 8
3y-1 = 8
3y = 8+1
3y = 9
divide both side by 3
y = 3

Akinola Ajayi
1 year ago

Let 2^y =a
a²-8=-2a
a²+2a-8=0
Solving using quadratic formula
a=2 or -4
Recall a=2^y
Therefore y = 1

Load More...