Solve for x log of x^2+ log of 25=2

$\log (x^{2}) + \log (25) = 2$

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (x^{2} \cdot 25) = 2$

Move $25$ to the left of $x^{2}$ .

$\log (25 x^{2}) = 2$

Rewrite $\log (25 x^{2}) = 2$ 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 $b^{y} = x$ .

$10^{2} = 25 x^{2}$

Solve for $x$ .

Rewrite the equation as $25 x^{2} = 10^{2}$ .

$25 x^{2} = 10^{2}$

Divide each term in $25 x^{2} = 10^{2}$ by $25$ and simplify.

Divide each term in $25 x^{2} = 10^{2}$ by $25$ .

$\frac{25 x^{2}}{25} = \frac{10^{2}}{25}$

Simplify the left side.

Cancel the common factor of $25$ .

Cancel the common factor.

$\frac{25 x^{2}}{25} = \frac{10^{2}}{25}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{10^{2}}{25}$

Simplify the right side.

Raise $10$ to the power of $2$ .

$x^{2} = \frac{100}{25}$

Divide $100$ by $25$ .

$x^{2} = 4$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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.

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.

$x = 2, - 2$

Solve for x log of x^2+ log of 25=2

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