Algebra 2 question - Exponential graphs?

Question

How would you find the equation of a graph that passes through the points (2,4) and (6,8) that follows the form of y=a*x^b?

Answer 1

(6/2)^b = 8/4
so b = ln 2 / ln 3
a 2^(ln2/ln3) = 4
so a = 4 / 2^(ln2/ln3)

Answer 2

rather what it boils suited right down to is which you plug interior the factors and you get 2 equations in 2 unknowns. 3 = ab^a million 12 = ab^2 the 1st equation tells us that a = 3/b. Plug that into the 2d to get 12 = (3/b)b^2 = 3b b = 4 a = 3/4 So the function is y = (3/4)4^x. an powerful trick for exponential applications is to take logarithms. enable's attempt this for the subsequent one. a million = ab^3 4 = ab^5 for each equation, take the log of each and every area and simplify. log(a million) = log(a) + 3*log(b) log(4) = log(a) + 5*log(b) we've a kit of linear equations that we are waiting to treatment for log(a) and log(b). Subtract the 1st equation from the 2d. log(4) - log(a million) = 2*log(b) log(4) = 2*log(b) (a million/2) log(4) = log(2) = log(b) b = 2 a = a million/8 y = (a million/8)2^x

Answer 3

Plug in the two points and you have
4 = a*(2^b)
8 = a*(6^b)

Solve both for a and set them equal to each other:
4 / (2^b) = 8 / (6^b)

This becomes
4(6^b) = 8(2^b)
(6^b) = 2(2^b)
6^b = 2^(b+1)

Now here's the key step: take the log of both sides to get the exponents down.
log (6^b) = log (2^(b+1))
b log(6) = (b+1) log(2)
b log(6) - b log(2) = log(2)
b( log(6) - log(2) ) = log(2)
b( log(3) ) = log(2)
b = log(2)/log(3)

We can get "a" by doing a substitution:
4 = a*(2^b)
a = 4 / 2^b
a = 2^2 / 2^b
a = 2^(2-b)
a = 2^(2 - log(2)/log(3))

So
y = 2^(2 - log(2)/log(3)) * x^[ log(2)/log(3) ]

I suppose you could play around with this and simplify it to get:
y = (2^2) [ 2^(-log(2)/log(3)) ] x^[ log(2)/log(3) ]
y = (2^2) [ (1/2)^(log(2)/log(3)) ] x^[ log(2)/log(3) ]
y = 4 [ x/2 ] ^(log[base 3] (2))

Answer 4

4=a*2^b
8=a*6^b
Dividing
2= 3^b so b= log2/log3=0.631
4= a*1.549 so a= 2.583