Algebra Question

[Kingdaniel101]

Can anyone confirm, or fix the working to this problem:

8^y x 4^(y^2-8) = 16 Solve for y?

Thanks in advance,

Daniel

 

[nycmathguy]

Can anyone confirm, or fix the working to this problem:

8^y x 4^(y^2-8) = 16 Solve for y?

Thanks in advance,

Daniel

8^y • 4^(y^2 - 8) = 16

2^[(3y) • 2^(2y^2 - 16)] = 2^(4)

Base 2 is the same on both sides of the equation. Bring down and equate exponents.

3y + 2y^2 - 16 = 4

3y + 2y^2 - 16 - 4 = 0

3y + 2y^2 - 20 = 0

2y^2 + 3y - 20 = 0

Can you take it from here?

I got y = - 4 and y = 5/2.

 

[MathLover1]

8^y * 4^(y^2-8) = 16 Solve for y

(2^3)^y * (2^2)^(y^2-8) =2^4

2^(3y) * 2^(2(y^2-8)) =2^4

2^(3y+2(y^2-8)) =2^4..........if base same, exponents are same to

3y+2(y^2-8) =4

3y+2y^2-16=4

2y^2+3y-16-4=0

2y^2+3y-20=0........factor

2y^2-5y+8y-20=0

(2y^2-5y)+(8y-20)=0

y(2y-5)+4(2y-5)=0

(y + 4) (2y - 5) = 0

=> y=-4
=>y=5/2

 

[nycmathguy]

8^y * 4^(y^2-8) = 16 Solve for y

(2^3)^y * (2^2)^(y^2-8) =2^4

2^(3y) * 2^(2(y^2-8)) =2^4

2^(3y+2(y^2-8)) =2^4..........if base same, exponents are same to

3y+2(y^2-8) =4

3y+2y^2-16=4

2y^2+3y-16-4=0

2y^2+3y-20=0........factor

2y^2-5y+8y-20=0

(2y^2-5y)+(8y-20)=0

y(2y-5)+4(2y-5)=0

(y + 4) (2y - 5) = 0

=> y=-4
=>y=5/2

I was right.