Please help...?

Question

Given that y = ax^b + 5, and that y = 7 when x = 3 and y = 9.5 when x = 2, find the value of a and of b.

Ans: a = 18, b = -2

Please help...

Answer 1

7 = a(3^b) + 5

2 = a(3^b)


then
9.5 = a(2^b) + 5

4.5 = a(2^b)

dividing...

2/4.5 = 3^b/2^b

(4/9) = (3/2)^b

ln(4/9) = b ln(3/2)
b = ln(4/9) / ln(3/2) = -2

or note ... (4/9) = (2/3)^2 = (3/2)^b ... thus b = -2.

2 = a(3^-2)
2 = a(1/9)
a = 18.


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Answer 2

Always keep this thought in mind: you will always need the same number of equations as there are unknowns. In this case, there are 2 unknowns, a & b.......this means that you will need 2 equations involving a & b in order to solve for both variables. The problem gives you a pair of x,y combinations that you can use for precisely this task:

Plug in the first (x,y) set, or (3,7), and write out the result:

7 = a*3^b + 5

a*3^b = 2

Solving for a gives us a = 2/3^b = 2*3^(-b)

Now, do the same thing for the other set, (2, 9.5):

9.5 = a*(2)^b + 5

a*2^b = 4.5

a = 4.5*2^(-b)

Set the 2 simplified equations for a equal to each other, then solve for b:

2*3^(-b) = 4.5*2^(-b)

1.5^(-b) = 2.25

At this point, since b is an exponent, the simple way to isolate it is to take the logarithm (or log) of each side.....

log (1.5)^(-b) = log 2.25

-b * log 1.5 = log 2.25

b = - (log 2.25/log 1.5) = -2

Now that we know b, we can find a:

a = 2*3^(-b) = 2*3^2 = 18 or

a = 4.5*2^(-b) = 4.5*2^2 = 4.5*4 = 18


Hope all of this helped you out a little.....

Answer 3

ok - we've got two unknowns and two equations.
substitute for x and y, gives:

7 = a*3^b + 5 => 2 = a * 3^b => 2/a = 3^b
=> log (2/a) = log(3^b)
=> log2 - log a = b log3
9.5 = a*2^b + 5 => 4.5 = a * 2^b => 4.5/a = 2^b
=> log (4.5/a) = log(2^b)
=> log4.5 - log a = b log 2

let A = log a

Now we have log
log2 - A = b log 3
log4.5 - A = b log 2
Regular algebraic substitution will work fine here:
A = log 2 - b log3
=> log 4.5 + b log3 - log 2 = b log 2
=> log (4.5 /2) = b log (2/3)
=> 4.5/2 = (2/3)^b
=> 9/4 = (2/3)^b
=> (3/2) ^2 = (2/3)^b
=> (2/3) ^-2 = (2/3) ^b
=> b = -2
or
b = log (9/4) / log (2/3)
=> b = -2
=> A = log 2 + 2 log3
=> A = log (2 * 3^2)
=> A = log 18
A = log a
=> a = 18.