Question: Find a and c in the equation

$$y=ca^x$$

when it is given that $x = \frac{1}{2}$ gives $y = 6$ and $x = 2$ gives $y = 162$

I already tried to solve when $6=ca^{\frac{1}{2}}$ and managed to get $a=\frac{36}{c^2}$ and when I insert $a$ in the equation $6=c\cdot \left(\frac{36}{c^2}\right)^{\frac{1}{2}}$, then I end up getting $0$ for $c$. Would appreciate help :)

Answer: 
$6 =ca^{\frac 12}$

$162 =ca^2$

Divide the second by the first (justified as $c \ne 0$),

$a^{\frac 32} = \frac{162}6 \implies a =(\frac{162}6)^{\frac 23} = $ (a nice round number you can work out).

Now put that into either to work out $c$.
Suppose we have $y=ca^{x}$. Then substituting the points $(x,y)=(\frac{1}{2},6)$ and $(x,y)=(2,162)$ gives $$6=ca^{\frac{1}{2}}$$
$$162=ca^{2}$$

Dividing the two we obtain $a^{\frac{3}{2}}=\frac{162}{6}=27$ so $a=9$. Now substituting into the first we obtain $$6=c(9)^{\frac{1}{2}}=3c$$

so $c=2$. Thus we have $y=2\cdot 9^{x}$.