What are the solutions to the equation (1+0.12)^1=(1+x)^{12} ?

The solutions to the equation (1+0.12)^1=(1+x)^{12} are x=0.00948…,x=-2.00948…

Find the zeros of (1+0.12)^1=(1+x)^{12}

The zeros of (1+0.12)^1=(1+x)^{12} are x=0.00948…,x=-2.00948…

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(1 + 0.12)^{1} = (1 + x)^{12}

x = 0.00948 \dots, x = - 2.00948 \dots

Solution steps

(1 + 0.12)^{1} = (1 + x)^{12}

Switch sides

(1 + x)^{12} = (1 + 0.12)^{1}

Simplify

(1 + 0.12)^{1} : 1.12

For

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

, n is even, the solutions are

Solve

The solutions are

x = 0.00948 \dots, x = - 2.00948 \dots

x + x^{3} = 2210

t^{3} - 6 t^{2} = 2 t

63 x - 2 x^{2} - x^{3} = 0

x^{5} = 1024

4 x^{5} - x^{3} - 4 x^{2} + 1 = 0

What are the solutions to the equation (1+0.12)^1=(1+x)^{12} ?
The solutions to the equation (1+0.12)^1=(1+x)^{12} are x=0.00948…,x=-2.00948…
Find the zeros of (1+0.12)^1=(1+x)^{12}
The zeros of (1+0.12)^1=(1+x)^{12} are x=0.00948…,x=-2.00948…