# Solve for x? if 4=(1+x)^24

##### 3 Answers
Jun 16, 2018

$- 1 + {2}^{\frac{1}{12}}$

#### Explanation:

$4 = {\left(1 + x\right)}^{24}$

$\sqrt[24]{4} = 1 + x$

${4}^{\frac{1}{24}} = 1 + x$

${2}^{\frac{2}{24}} = 1 + x$

${2}^{\frac{1}{12}} = 1 + x$

$- 1 + {2}^{\frac{1}{12}} = x$

Jun 17, 2018

Extend to complex numbers:

If anyone has studies complex numbers

#### Explanation:

$4 = {\left(1 + x\right)}^{24}$

$4 = {\left(1 + x\right)}^{24} {e}^{2 k \pi i}$

as ${e}^{2 k \pi i} = 1 , \forall k \in \mathbb{Z}$

${4}^{\frac{1}{24}} = \left(1 + x\right) {e}^{\frac{1}{12} k \pi i}$

$\implies {2}^{\frac{1}{12}} = {e}^{\frac{1}{12} k \pi i} + x {e}^{\frac{1}{12} k \pi i}$

$\implies {2}^{\frac{1}{12}} - {e}^{\frac{1}{12} k \pi i} = x {e}^{\frac{1}{12} k \pi i}$

$\implies x = \frac{{2}^{\frac{1}{12}} - {e}^{\frac{1}{12} k \pi i}}{e} ^ \left(\frac{1}{12} k \pi i\right)$

$\implies k = \left\{0 , 1 , 2 , 3 , \ldots , 22 , 23\right\}$

Jun 30, 2018

$x = {2}^{\frac{1}{12}} - 1$

#### Explanation:

We can take the $24$th root of both sides to get

${4}^{\frac{1}{24}} = 1 + x$

Subtracting $1$ from both sides gives us

$x = {4}^{\frac{1}{24}} - 1$

We can now rewrite $4$ as ${2}^{2}$. This gives us

$x = {2}^{2 \cdot \frac{1}{24}} - 1$

which can be simplified as

$x = {2}^{\frac{1}{12}} - 1$

Hope this helps!