How do you find the exponential function that contains both of these points (2,12.6) and (5, 42.525)?

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Answer 1 · 2015-06-18T23:40:48

Suppose $f(x) = k*a^x$ for some $k in RR$ and $a > 0$ , $a != 1$

Then $a = root(3)(42.525/12.6) = root(3)(3.375) = 1.5$

and $k = 12.6/(a^2) = 12.6/2.25 = 5.6$

So $f(x) = 5.6(1.5)^x$

Suppose $f(x) = k*a^x$ for some $k in RR$ and $a > 0$ , $a != 1$

Then $f(5)/f(2) = (k*a^5)/(k*a^2) = a^3$

So $a = root(3)(f(5)/f(2)) = root(3)(42.525/12.6) = root(3)(3.375) = 1.5$

$f(2) = k*a^2$ , so $k = f(2)/(a^2) = 12.6/(1.5^2) = 12.6 / 2.25 = 5.6$

So $f(x) = 5.6(1.5)^x$