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

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3+ 2sqrt2)^x + (3-2sqrt2)^x = 34

We know thta:

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

Now substitute:

==> (1/(3-2sqrt2)^x + (3-2sqrt2)^x = 34

==> (3-2sqrt2)^-x + (3-2sqrt2)^x = 34

Let y= (3-2sqrt3)^x

==> 1/y + y = 34

Multily by y:

==> 1+y^2 = 34y

==> y^2 - 34y +1 = 0

==> y1= [34 + sqrt(1152)]/ 2 = 17 + 12sqrt2

==> 17 + 12sqrt2 = (3-2sqrt2)^x1

==> x1= log (3-2sqrt2) (17+12sqrt2)

==> y2= 17-12sqrt2

==> x2= log (3-2sqrt2) (17-12sqrt2)

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)

(3+2sqrt2)^x +(3-2sqrtx)^x = 34.To solve for x.

Solution:

So we can write the 2nd term as : 3-2sqrtx = [3^2-(2sqrt2)^2]/(3^2+2sqrt2). = (9-8)/(3+2sqrt2) = 1/(3+2sqrt2).

Therefore the given equation becomes:

a +(1/a) = 34. where a = (3+2sqrt2)^x

a+1/a = 34

a^2+1 = 34a

a^2-34a+1 = 0

a = [34+sqrt(34^2-4)/2 or {34+ sqrt(34^2-4)}/2

a = 17 + sqrt 288= 17 + 6 sqrt 2

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

We see that (3+2sqrt2)^2 = 3^2+4*2+ 12sqrt2 = 17 +12sqrt.

Therefore x = 2.

Similarly taking a = (3+2sqrt2)^x = 17 -12sqart2 we get:

( 3+2sqrt)^(-2) = 1/(3+2sqrt2)^2 = 1/(17+12sqrt2) = 17-12sqrt2)/[17^2-12sqrt)^2] = (17-12sqrt2)/(289-288)) = 17-12sqrt2.

So x=-2 .

Therefore x =2 or x= -2.