What are the solutions to the equation 4=(1+r/2)^{20} ?

The solutions to the equation 4=(1+r/2)^{20} are r=2\sqrt[10]{2}-2,r=-2\sqrt[10]{2}-2

Find the zeros of 4=(1+r/2)^{20}

The zeros of 4=(1+r/2)^{20} are r=2\sqrt[10]{2}-2,r=-2\sqrt[10]{2}-2

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4 = (1 + \frac{r}{2})^{20}

Decimal Notation

r = 0.14354 \dots, r = - 4.14354 \dots

Solution steps

4 = (1 + \frac{r}{2})^{20}

Switch sides

(1 + \frac{r}{2})^{20} = 4

For

(g (x))^{n} = f (a)

, n is even, the solutions are

Solve

The solutions are

9 x^{3} - 4 = 0

X^{3} - 64 X = 0

b^{4} = 339

x^{10} = 1000

r^{4} = \frac{16}{81}

What are the solutions to the equation 4=(1+r/2)^{20} ?
The solutions to the equation 4=(1+r/2)^{20} are r=2\sqrt[10]{2}-2,r=-2\sqrt[10]{2}-2
Find the zeros of 4=(1+r/2)^{20}
The zeros of 4=(1+r/2)^{20} are r=2\sqrt[10]{2}-2,r=-2\sqrt[10]{2}-2