[tex]f(x) = ae {}^{bx} \\ f(2) = 5 \: \: \: \: \: f(3) = 4[/tex]
[tex]5 = ae {}^{2b} \: \: \: \: \: \: \: 4 = ae {}^{3b} \\ divide \: both \: systems \\ e {}^{b} = \frac{4}{5} \: \: \: \: so \: \: \: b = ln(0.8) \\ solving \: for \: a \: we \: get \: that \: a = \frac{125}{16} [/tex]
[tex]f(x) = \frac{125}{16} e {}^{ln(0.8)x} [/tex]
[tex]f(10) = \frac{125}{16} e {}^{10ln(0.8)} = \frac{65536}{78125} \approx0.838861 \\ f(100) = \frac{125}{16} e {}^{100ln(0.8)} = \frac{4 {}^{98} }{5 {}^{97} } \approx1.59 \times 10 {}^{ - 9} [/tex]
