f(x) is an exponential function, so [tex]f(x) = a\cdot b^x[/tex].
f(11) = 28 tells us [tex]28 = a\cdot b^{11}[/tex]
f(5) = 3 tells us [tex]3 = a\cdot b^5[/tex]
If we divide those equations, we'd have
[tex]\dfrac{28}{3} = \dfrac{a\cdot b^{11}}{a \cdot b^5}[/tex]
This simplifies to
[tex]\dfrac{28}{3} = b^6[/tex]
Take the sixth root of both sides and you have
[tex]\sqrt[6]{\dfrac{28}{3}} = b[/tex]
(technically it is ±, but b>0 for exponential functions)
Now using that, we have [tex]f(x) = a\cdot \left(\sqrt[6]{\frac{28}{3}}\right)^x[/tex].
Take a deep breath...
OK, we plug in 5 and 3 one more time to find a.
[tex]3= a\cdot \left(\sqrt[6]{\frac{28}{3}}\right)^5[/tex]
Divide by that mess on the left and you'll have a.
[tex]\dfrac{3}{\left(\sqrt[6]{\frac{28}{3}}\right)^5}= a[/tex]
And out function is [tex]f(x) = \dfrac{3}{\left(\sqrt[6]{\frac{28}{3}}\right)^5} \cdot \left(\sqrt[6]{\frac{28}{3}}\right)^x[/tex].
Wow, that's ugly.
You could use exponent rules to simplify this a bit:
[tex]f(x) =3\cdot \left(\sqrt[6]{\frac{28}{3}}\right)^{x-5}[/tex]
Now evaluate f(8) by putting 8 in for x and you'll get 9.16515138991, or 9.17 to the nearest hundredth.
