Solve the equation (3+2sqrt2)^x + (3-2sqrt2)^x=34

The given equation is an exponential equation and we'll
solve it using the substitution technique.

We notice that
(3+2sqrt2)*(3-2sqrt2) = 3^2 -
(2sqr2)^2

(3+2sqrt2)*(3-2sqrt2) = 9 -
4*2

(3+2sqrt2)*(3-2sqrt2) = 9 -
8

(3+2sqrt2)*(3-2sqrt2) =
1

We'll raise to x power both
sides:

[(3+2sqrt2)*(3-2sqrt2)]^x =
1^x

(3+2sqrt2)^x*(3-2sqrt2)^x =
1

(3+2sqrt2)^x = 1 /
(3-2sqrt2)^x

We'll note (3+2sqrt2)^x =
t

t = 1 /
(3-2sqrt2)^x

(3-2sqrt2)^x =
1/t

We'll substitute in the given
equation:

t + 1/t = 34

t^2 + 1
- 34t = 0

t^2  - 34t + 1 =
0

We'll apply the quadratic
formula:

t1 =
(34+sqrt1152)/2

t1 =
(34+24sqrt2)/2

t1 = 17 +
12sqrt2

t2 = 17 -
12sqrt2

(3+2sqrt2)^x = 17 +
12sqrt2

lg (3+2sqrt2)^x = lg (17 +
12sqrt2)

x*lg (3+2sqrt2) = lg (17 +
12sqrt2)

x1 = lg (17 + 12sqrt2) / lg
(3+2sqrt2)

x2 = lg (17 - 12sqrt2) / lg
(3+2sqrt2)