What are the solutions to the equation (1+0.02)^{12}=(1+x)^4 ?

The solutions to the equation (1+0.02)^{12}=(1+x)^4 are x=0.061208,x=-2.061208

Find the zeros of (1+0.02)^{12}=(1+x)^4

The zeros of (1+0.02)^{12}=(1+x)^4 are x=0.061208,x=-2.061208

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(1 + 0.02)^{12} = (1 + x)^{4}

x = 0.061208, x = - 2.061208

Solution steps

(1 + 0.02)^{12} = (1 + x)^{4}

Switch sides

(1 + x)^{4} = (1 + 0.02)^{12}

Simplify

(1 + 0.02)^{12} : 1.26824 \dots

For

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

, n is even, the solutions are

Solve

The solutions are

x = 0.061208, x = - 2.061208

m (m - 1) (m - 2) (m - 3) = 840

2 x^{3} + 4 x = 6 x^{2}

5 y^{3} - 44 y^{2} - 9 y = 0

x^{3} = - 3 x^{2} + 3 x - 1

- 3 x^{4} = 0

What are the solutions to the equation (1+0.02)^{12}=(1+x)^4 ?
The solutions to the equation (1+0.02)^{12}=(1+x)^4 are x=0.061208,x=-2.061208
Find the zeros of (1+0.02)^{12}=(1+x)^4
The zeros of (1+0.02)^{12}=(1+x)^4 are x=0.061208,x=-2.061208