# Solve inequality (7+4`sqrt3` )^x+(7-4`sqrt3` )^x =< 4

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may use the following alternative approach, such that:

`(7 + 4sqrt3)(7 - 4sqrt3) = 49 - 48 = 1 => (7 + 4sqrt3) = 1/(7 - 4sqrt3) => (7 + 4sqrt3)^x = 1/(7 - 4sqrt3)^x`

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

`(7 + 4sqrt3)^x = y => 1/(7 - 4sqrt3)^x = 1/y`

Replacing the variable, yields:

`y + 1/y <= 4 => y^2 - 4y + 1 <= 0`

You need to attach the quadratic equation, such that:

`y^2 - 4y + 1 = 0`

`y_(1,2) = (4+-sqrt(16 - 4))/2 => y_(1,2) = (4+-sqrt12)/2`

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

You should notice that the inequality holds for `y in [2-sqrt3,2+sqrt3]`

You need to evaluate x, such that:

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

Taking common logarithms yields:

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

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

Hence, the inequality holds if `x in [(ln (2 - sqrt3))/(ln (7 + 4sqrt3)), (ln (2 + sqrt3))/(ln (7 + 4sqrt3))].`

aruv | High School Teacher | (Level 2) Valedictorian

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`(7+4sqrt(3))=4+3+2xx2xxsqrt(3)=(2+sqrt(3))^2`

`and`

`(7-4sqrt(3))=4+3-2xx2xxsqrt(3)=(2-sqrt(3))^2`

Thus given problem can be re written as

`((2+sqrt(3))^2)^x+((2-sqrt(3))^2)^x<=4`

`(2+sqrt(3))^(2x)+(2-sqrt(3))^(2x)<=4`

Define a function

`f(x)=4-(2+sqrt(3))^(2x)-(2-sqrt(3))^(2x)`

we want to find x for which  `f(x)>=0`

So draw a graph of the function f(x)

Thus `f(x)>=0`  if `x in[-1/2,1/2]`