# solve ((4-(15)^0.5)^0.5)^x+((4+(15)^0.5)^0.5)^x=8?

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### 1 Answer

You need to solve the exponential equation `(sqrt(4-sqrt15))^x + (sqrt(4+sqrt15))^x =8` , hence you should notice that `4+sqrt15 ` is conjugate to `4-sqrt15` .

You should perform the following test such that:

`sqrt (4+sqrt15) *sqrt (4-sqrt15) = sqrt ((4+sqrt15) *(4-sqrt15))`

`sqrt ((4+sqrt15) *(4-sqrt15)) = sqrt (16-15) = sqrt 1 = 1`

Hence, `sqrt (4+sqrt15) = 1/sqrt (4-sqrt15) `

Raising both sides to x power yields:

`(sqrt (4+sqrt15))^x = 1/((sqrt (4-sqrt15) )^x)`

You should come up with the substitution such that:

`(sqrt (4+sqrt15))^x = y =gt ((sqrt (4-sqrt15) )^x) = 1/y`

Substituting y for `(sqrt (4+sqrt15))^x` and `1/y` for `((sqrt (4-sqrt15) )^x)` yields:

`y + 1/y = 8`

You need to bring all terms to a common denominator such that:

`y^2 + 1 = 8y`

You need to move all terms to the left side such that:

`y^2 - 8y + 1 = 0`

You should use quadratic formula such that:

`y_(1,2)= (8 +- sqrt(8^2 - 4*1*1))/2`

`y_(1,2)= (8 +- sqrt(60))/2`

`y_(1,2)= (8 +- 2sqrt(15))/2`

`y_(1,2)= (4 +- sqrt(15))`

You need to solve for x the equations `(sqrt (4+sqrt15))^x = (4 + sqrt(15)) ` and `(sqrt (4+sqrt15))^x =(4+ sqrt(15))^(-1)` such that:

`(sqrt (4+sqrt15))^x = (4 + sqrt(15))`

`(4+sqrt15)^(x/2) = (4 + sqrt(15)) =gt x/2 = 1 =gt x = 2`

`(sqrt (4+sqrt15))^x = (4 + sqrt(15))^(-1)`

`(4+sqrt15)^(x/2) = (4 + sqrt(15))^(-1) =gt x = -2`

**Hence, evaluating solutions to exponential equation yields x = 2 and x = -2.**