Logarithms....................

I need help on this because i dont get what the question is asking for. Help needed.

Given that y= ax^b + 7, that y = 79 when x=2 and that y=16 when x = 4, calculate the numerical values of a and of b.

y= ax^b + 7
and 79 = a* 2^b + 7
=> 72/(2^b) = a
and 16 = a *4^b + 7
=> 9/(4^b) = a

=> 72/(2^b) = 9/(4^b)
=> 72 * 4^b = 9 * 2^b
=> 72 * 2^(2b) = 9 * 2^b
=> log₂(72 * 2^(2b)) = log₂(9 * 2^b)
=> log₂(72) + log₂(2^(2b)) = log₂(9) + log(2^b)
=> log₂(72) + 2b = log₂(9) + b
=> b = log₂(9) - log₂(72)
=> b = log₂(9/72)
=> b = log₂(1/8)
=> b = log₂(2^(-3))
=> b = -3

now 72/(2^(-3)) = a
=> 72 * 8 = a
=> a = 576

79 = a*2^b+7 >>>> 72 = a*2^b
16 = a*4^b+7 >>>> 9 = a*2^(2b)
Divide equations
8 = 2^(-b) = 2^3
b = -3
72 = a*2^-3 = a/8
a = 576
b = -3