Given:
f(x) is an exponential function where f(3) = 18 and f(7.5) = 60.
To find:
The value of f(12), to the nearest hundredth.
Solution:
Let the exponential function be
[tex]f(x)=ab^x[/tex] ...(i)
It is given that [tex]f(3) = 18[/tex]. Substitute [tex]x=3[/tex] in (i).
[tex]f(3)=ab^3[/tex]
[tex]18=ab^3[/tex] ...(ii)
It is given that [tex]f(7.5) = 60[/tex]. Substitute [tex]x=7.5[/tex] in (i).
[tex]f(7.5)=ab^{7.5}[/tex]
[tex]60=ab^{7.5}[/tex] ...(iii)
Divide (iii) by (ii).
[tex]\dfrac{60}{18}=\dfrac{ab^{7.5}}{ab^3}[/tex]
[tex]\dfrac{10}{3}=b^{4.5}[/tex]
[tex]\left(\dfrac{10}{3}\right)^\frac{1}{4.5}=b[/tex]
[tex]b\approx 1.30676[/tex]
Putting [tex]b=1.30676[/tex] in (ii), we get
[tex]18=a(1.30676)^3[/tex]
[tex]\dfrac{18}{(1.30676)^3}=a[/tex]
[tex]a\approx 8.0665[/tex]
Putting [tex]a=8.0665[/tex] and [tex]b=1.30676[/tex] in (i), we get
[tex]f(x)=8.0665(1.30676)^x[/tex]
Putting x=12, we get
[tex]f(12)=8.0665(1.30676)^{12}[/tex]
[tex]f(12)=200.0024[/tex]
[tex]f(12)\approx 200.00[/tex]
Therefore, the value of f(12) is about 200.00.