Answer
$\displaystyle \qquad f(x)=\frac{10}{1+9(1.5^{-x})}$
Work Step by Step
Goal: $\displaystyle \quad f(x)=\frac{N}{1+Ab^{-x}}$,
where $N$ = limiting value = $10$ (given)
$f(0)=1\Rightarrow\left\{\begin{array}{ll}
1 & =\frac{10}{1+Ab^{0}}\\
& \\
1+A & =10 \\
A & =10-1=9
\end{array}\right.\qquad\Rightarrow\qquad A=9.$
For small values of x, $f(x)=\displaystyle \frac{N}{1+Ab^{-x}}\approx(\frac{N}{1+A})b^{x}$
(the logistic function grows approximately exponentially with base b.)
$f(x)\displaystyle \approx(\frac{10}{10})b^{x}=b^{x}$
To increase by $ 50\%$ with an increase of x by 1, means that
$f(x+1)=1.50f(x)$
$ b^{x+1}=1.5( b^{x})\qquad$ (divide with $ b^{x})$
$b=1.5$
Thus, $\displaystyle \qquad f(x)=\frac{10}{1+9(1.5^{-x})}$
