Solve the equation (square root 5)^(2+4+...+2x) = 5^45 , if x is a natural number.

We'll re-write the left side
term:

(sqrt 5)^(2+4+...+2x) =
5^[(2+4+...+2x)/2]

We'll factorize by 2 within brackets
from superscript:

5^[(2+4+...+2x)/2] = 5^[2*(1+2+...+2x)/2]
=> 5^[(1+2+...+x)]

We'll re-write the
equation:

5^[(1+2+...+x)] =
5^45

Since the bases are matching, we'll apply one to one
rule:

1+2+...+x = 45

We notice
that the sum from the left is the sum of the terms of an arithmetical
sequence:

1 + 2 + ... + x =
x*(1+x)/2

The equation will
become:

x*(1+x)/2 = 45

x^2 + x
= 90

x^2 + x - 90 = 0

We'll
apply quadratic formula:

x1 = [-1+sqrt(1 +
360)]/2

x1 = (-1+sqrt361)/2

x1
= (-1+19)/2

x1 = 9

x2 =
-10

Since the value of x has to be natural, w'ell reject
the negative value and we'll keep as soution of equation x = 9.