# Solve equation (9-2^x)^1/3 +(2^x+7)^1/3=4? x=??

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### 1 Answer

You should raise to cube both sides such that:

`(root(3)(9 - 2^x) + root(3)(2^x + 7))^3 = 4^3`

You need to expand the cube such that:

`(root(3)(9 - 2^x) + root(3)(2^x + 7))^3 = (root(3)(9 - 2^x))^3 + (root(3)(2^x + 7))^3 + 3root(3)((9 - 2^x)(2^x + 7))(root(3)(9 - 2^x) + root(3)(2^x + 7))`

You should substitute `4^3` for `root(3)(9 - 2^x) + root(3)(2^x + 7)` such that:

`9 - 2^x + 2^x + 7 + 3*64*root(3)((9 - 2^x)(2^x + 7)) = 64`

Eliminating like terms yields:

`192root(3)((9 - 2^x)(2^x + 7)) = 64 - 16`

`192root(3)((9 - 2^x)(2^x + 7)) = 48`

You need to divide by 48 both sides such that:

`4root(3)((9 - 2^x)(2^x + 7)) = 1 =gt root(3)((9 - 2^x)(2^x + 7)) = 1/4`

You need to raise to cube to remove the cube root such that:

`((9 - 2^x)(2^x + 7)) = 1/64`

`9*2^x + 63 - 2^(2x) - 7*2^x = 1/64`

`- 2^(2x) + 2*2^x + 63 - 1/64 = 0`

`-64*2^(2x) + 128*2^x + 4031 = 0`

You should come up with the substitution `2^x = y` such that:

`64y^2 - 128y - 4031 = 0`

You should use quadratic formula such that:

`y_(1,2)= (128 +- sqrt(1048320))/128`

`y_(1,2) = (128+-48sqrt455)/128`

You should solve for x `2^x = y` such that:

`2^x = (128+48sqrt455)/128 =gt x = ln((128+48sqrt455)/128)/ln 2`

`x~~3.170`

Since `(128-48sqrt455)/128 lt 0 =gt 2^x = (128+48sqrt455)/128 ` is not possible to be solved since `2^x gt 0` .

**Hence, evaluating the solution to the given equation yields `x~~3.170.` **