How do I find the equation of an exponential function that passes through (1,6) and (3,24)?

1 Answer
Jul 14, 2015

Answer:

Let #f(x) = ab^x# where #a != 0# and #b > 0#.

Use known values #f(1) = 6# and #f(3) = 24# and solve to find #a=3#, #b=2#

Hence #f(x) = 3*2^x#

Explanation:

The standard formula for an exponential function is:

#f(x) = ab^x#, with #a != 0# and #b > 0#

Given: #f(1) = 6# and #f(3) = 24#

#4 = 24/6 = f(3) / f(1) = (ab^3)/(ab^1) = b^2#

So #b = +-sqrt(4) = +-2#.

We want #b > 0#, so pick #b = 2#.

Then #a = (ab^1)/b = f(1)/b = 6 / 2 = 3#

So #a=3# and #b=2#, giving us #f(x) = 3 * 2^x#

General case:

Given #f(x_1) = y_1# and #f(x_2) = y_2#

#b^(x_2 - x_1) = (ab^(x_2)) / (ab^(x_1)) = f(x_2)/f(x_1)#

So #b = (f(x_2) / f(x_1))^(1/(x_2 - x_1))#

Then #a = f(x_1) / (b^(x_1))#