What is solution of equation (root(2-root3))^x+(root(2+root3))^x=4?

1 Answer | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to solve for x the following equation such that:

`(sqrt(2-sqrt3))^x + (sqrt(2+sqrt3))^x = 4`

Using the properties of exponentials yields:

`(2-sqrt3)^(x/2) + (2+sqrt3)^(x/2) = 4`

Since the radicands `2-sqrt3` and `2+sqrt3` are irrational conjugates, you should consider the use of the following relation, such that:

`(2+sqrt3) = 1/(2-sqrt3)`

Raising to the power `x/2` yields:

`(2+sqrt3)^x= 1/(2-sqrt3)^x`

You should come up with the following substitution, such that:

`(2+sqrt3)^x = y => 1/(2-sqrt3)^x = 1/y`

Replacing `y` and `1/y` for `(2+sqrt3)^x` and `1/(2-sqrt3)^x` yields:

`sqrt y + sqrt(1/y) = 4`

Squaring both sides yields:

`(sqrt y + sqrt(1/y))^2 = 4^2`

`y + 2sqrt(y*(1/y)) + 1/y = 16 => y + 2 + 1/y = 16`

`y^2 + 2y + 1 = 16y => y^2 + 2y + 1 - 16y = 0`

`y^2 - 14y + 1 = 0`

Using quadratic formula yields:

`y_(1,2) = (14+-sqrt(14^2 - 4))/2 => y_(1,2) = (14+-sqrt(192))/2`

`y_(1,2) = (14+-8sqrt3)/2 => y_(1,2) = (7+-4sqrt3)`

Replacing back `(2+sqrt3)^x` for y yields:

`(2+sqrt3)^x = (7+4sqrt3) => x*ln (2+sqrt3) = ln (7+4sqrt3) `

`x = (ln (7+4sqrt3))/(ln (2+sqrt3))`

`(2+sqrt3)^x = (7-4sqrt3) => x*ln (2+sqrt3) = ln (7-4sqrt3)`

`x = (ln (7-4sqrt3))/(ln (2+sqrt3))`

Hence, evaluating the solutions to the given equation yields `x = (ln (7+4sqrt3))/(ln (2+sqrt3))` and `x = (ln (7-4sqrt3))/(ln (2+sqrt3)).`

We’ve answered 318,996 questions. We can answer yours, too.

Ask a question