+26367 5^2*5^4*5^6*.......5^(2n) = (0.008)^(-30) find the value of n
\(\begin{array}{|rclrcl|} \hline 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 0.008^{(-30)} \\ && 0.008 &=& \frac{8}{1000} \\ && &=& \frac{2^3}{10^3} \\ && &=& \left( \frac{2}{10} \right)^3 \\ && &=& \left( \frac{1}{5} \right)^3 \\ && &=& \frac{1}{5^3} \\ && &=& 5^{-3} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& (5^{-3})^{-30} \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{(-3)\cdot (-30) } \\ 5^2\cdot 5^4 \cdot 5^6 \cdot \ldots \cdot 5^{2n} &=& 5^{90} \\ 5^{2+4+6 \cdot \ldots \cdot 2n } &=& 5^{90} \\\\ 2+4+6 \cdot \ldots \cdot 2n &=& 90 \\ 2\cdot(1+2+3 + \ldots + n ) &=& 90 \\ && 1+2+3 + \ldots + n &=& \frac{(1+n)\cdot n}{2} \\ 2\cdot \Big(\frac{(1+n)\cdot n}{2} \Big) &=& 90 \\ (1+n)\cdot n &=& 90 \\ n^2+n -90 &=& 0 \\ (n-9)\cdot(n+10) &=& 0 \\ \hline \end{array}\)
\(n = 9 \text{ or } n = -10\)
Because n > 0:
n = 9
\(5^2\cdot 5^4 \cdot 5^6\cdot 5^8\cdot 5^{10}\cdot 5^{12} \cdot 5^{14}\cdot 5^{16}\cdot 5^{18} = 0.008^{(-30)} \)
