Too Hard?

Let's find $f(10)$ first,

We use $x=5$ to find the value of $f(10)$

$f(2\times 5 - 3) = \frac{f(2\times 5) + 1}{f(2\times 5)-1}$

$f(7) = \frac{f(10) +1}{f(10) - 1}$

$5 = \frac{f(10) +1}{f(10) - 1}$

$5(f(10)-1) = f(10) +1$

$5f(10) - 5 = f(10) +1$

$4f(10)= 6$

$f(10) = \frac{3}{2}$

Now let's find $f(4)$ ,

We use $x=3\frac{1}{2}$ to find the value of $f(4)$

$f(2\times 3\frac{1}{2} - 3) = \frac{f(2\times 3\frac{1}{2}) + 1}{f(2\times 3\frac{1}{2})-1}$

$f(4) = \frac{f(7) +1}{f(7) - 1}$

$f(4) = \frac{5 +1}{5-1}$

$f(4) = \frac{3}{2}$

Now find $\frac { f(10)}{f(4)}$

$\frac { f(10)}{f(4)}= \frac{\frac{3}{2}}{\frac{3}{2}} = \boxed{1}$

$f(2x-3)=\dfrac {f(2x)+1}{f(2x)-1} -------- (1)$ $f(2x-3)f(2x)-f(2x-3)=f(2x)+1$ $f(2x)=\dfrac {f(2x-3)+1}{f(2x-3)-1}$ Put $2x \rightarrow 2x+3$ : $f(2x+3)=\dfrac {f(2x)+1}{f(2x)-1}=f(2x-3)$ from equation (1).

Thus $f(10)=f(4)$ and $\dfrac {f(10)}{f(4)}$ $=\dfrac {f(7+3)}{f(7-3)}$ $=\boxed{1}$