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

$\log (x) + \log (x) = \log (25)$

Simplify the left side.

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

$\log (x \cdot x) = \log (25)$

Multiply $x$ by $x$ .

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

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$x^{2} = 25$

Solve for $x$ .

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

$x = \pm \sqrt{25}$

Simplify $\pm \sqrt{25}$ .

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

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

$x = \pm 5$

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.

$x = 5$

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

$x = - 5$

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

$x = 5, - 5$

Exclude the solutions that do not make $\log (x) + \log (x) = \log (25)$ true.

$x = 5$