equations simultaneous non-linear

y  = pq^x

Note that when  x  = 0,   then y  = 5

This suggests that

5 = pq^0

5  = p (1)

5  = p

And the point  (4, 405)  is on the graph..so

405  = 5q^(4)    divide both sides by 5

81  = q^4        take the 4th root of each side and we get that

3  = q

So....to find  k   when  x  =2, we have

k  =  5*(3)^2   =  5 * 9   =  45

 equations simultaneous non-linear

geometric sequence: \(\text{$a_1 = p$, $a_2 = pq$, $a_3 = pq^2$, $a_4=pq^3$, $a_5 = pq^4$, $\ldots$}\)

\(\text{ $k \quad(a_3=pq^2)$ is the geometric mean of $y = 5 \quad(a_1=p)$ and $y = 405 \quad(a_5=pq^4)$ } \\ \huge{ \begin{array}{|rcll|} \hline k&=&\sqrt{5\cdot 405} \\ &=&\sqrt{2025} \\ &=& 45 \\ \hline \end{array} }\)