What is the answer to solvefor n,\sqrt[n]{x}=x^{n+2} ?

The answer to solvefor n,\sqrt[n]{x}=x^{n+2} is n=-1+sqrt(2),n=-1-sqrt(2)

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solvefor n,\sqrt[n]{x}=x^{n+2}

solve for

n = - 1 + 2, n = - 1 - 2

Decimal Notation

n = 0.41421 \dots, n = - 2.41421 \dots

Solution steps

Apply exponent rules

\frac{1}{n} = n + 2

Solve

\frac{1}{n} = n + 2 : n = - 1 + 2, n = - 1 - 2

n = - 1 + 2, n = - 1 - 2

7^{6 x + 7} = 2

x + 5 = 5^{x}

2^{3 x - 1} = \frac{1}{2}

2 e^{2 x} + 5 e^{x} - 3 = 0

x, x y = e^{4 x - y} - 1

What is the answer to solvefor n,\sqrt[n]{x}=x^{n+2} ?
The answer to solvefor n,\sqrt[n]{x}=x^{n+2} is n=-1+sqrt(2),n=-1-sqrt(2)