Logarithm Question URGENT HELP?

Question

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

Answer 1

79 = a x 2^b + 7
16 = a x 4^b + 7

72 = a x 2^b
9 = a x 4^b

72 = a x 2^b
9 = a x 2^(2b)
8 = 2^(-b)
2^b = 1/8
b log 2 = log(1/8) = log 2^(-3) = - 3.log 2
b = - 3
9 = a x 2^(-6)
a = 9 x 2^6
a = 576
ANSWER
a = 576
b = - 3

Answer 2

79 = a*2^b + 7
16 = a*4^b + 7

72 = a*2^b
9 = a*4^b

Divide equations to get:

8 = 1/(2^b)
b = -3

So: 72 = a/8
a = 576

Answer 3

79 = a*( 2^b ) + 7
This means:

1) a*( 2^b ) = 72


16 = a*( 4^b ) + 7
This means

2) a*[ 2^( 2b ) ] = 9

Now divide left and Wright side of 2) by those of 1):

a*[ 2^( 2b ) ]/a*( 2^b ) = 9/72

This gives:

2^( 2b-b ) = 1/8

Then

2^b = 2^(-3)

b = -3

To find a you may substitute b by -3 in 1)

a*( b^(-3) ) = 72

a = 8*72 = 576

Answer 4

79 = a. 2^b + 7
16 = a. 4^b + 7

72 = a.2^b
9 = a.4^b

8 = 2^b / 4^b

8 = (2/4)^b .......................... but 8 = 2^3 = (½)^ -3

(½) ^ -3 = (½) ^ b

b = -3 and a = 576

Answer 5

when y=79 and x=2,
a2^b=72
a= 72/2^b (eq. 1)

when y=16 and x=4,
a4^b=9
a= 9/4^b (eq. 2)

subs. eq. 1 and eq.2:
72/2^b=9/4^b
4^b/2^b=9/72
2^b=1/8
2^b=2^-3
b= -3
put b= -3 in eq. 1
a=72/2^-3
a=72*8
a=576

a=576, b= -3