+0

# equations simultaneous non-linear

+1
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lynx7  Apr 13, 2018
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#1
+85726
+3

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

CPhill  Apr 13, 2018
#2
+19207
+1

equations simultaneous non-linear

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

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

heureka  Apr 16, 2018
edited by heureka  Apr 16, 2018
edited by heureka  Apr 16, 2018

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