What are the solutions to the equation (1+x)^{20}=40 ?

The solutions to the equation (1+x)^{20}=40 are x=\sqrt[20]{40}-1,x=-\sqrt[20]{40}-1

Find the zeros of (1+x)^{20}=40

The zeros of (1+x)^{20}=40 are x=\sqrt[20]{40}-1,x=-\sqrt[20]{40}-1

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(1 + x)^{20} = 40

Decimal Notation

x = 0.20254 \dots, x = - 2.20254 \dots

Solution steps

(1 + x)^{20} = 40

For

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

, n is even, the solutions are

Solve

The solutions are

x^{2} \cdot x^{2} = 81

(1 - x)^{10} = 0.75

m^{1} m^{2} = - 1

3 x^{4} - 15 x^{2} + 29 = 0

a^{3} - 3 a^{2} = 240

What are the solutions to the equation (1+x)^{20}=40 ?
The solutions to the equation (1+x)^{20}=40 are x=\sqrt[20]{40}-1,x=-\sqrt[20]{40}-1
Find the zeros of (1+x)^{20}=40
The zeros of (1+x)^{20}=40 are x=\sqrt[20]{40}-1,x=-\sqrt[20]{40}-1