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

### 2 Answers | Add Yours

We have to solve (sqrt 5)^(2+4+...+2x) = 5^45

(sqrt 5)^(2+4+...+2x) = 5^45

=> (5^(1/2))^(2+4+...+2x) = 5^45

=> 5^(0.5*(2+4+...+2x)) = 5^45

as the base is the same equate the exponent

0.5*(2+4+...+2x) = 45

=> 2 + 4 + 6 + 8 ...+ 2x = 90

=> 2 + 4 + 6 + 8 + ...+ 18 = 90

=> 2x = 18

=> x = 9

**The required value of x = 9**

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.