Solve for x log base x of 4=2

$\log_{x} (4) = 2$

Rewrite $\log_{x} (4) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b$ $\neq$ $1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$x^{2} = 4$

Solve for $x$

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Take the $(+2)th$ root of both sides of the $equation$ to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Exclude the solutions that do not make $\log_{x} (4) = 2$ true.

$x = 2$

Solve for x log base x of 4=2

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