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If f(x) is an exponential function where f(5) = 3 and f(11) = 28, then find the
value of f(8), to the nearest hundredth.


Sagot :

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.