Solve for x log base 3 of 3x^2+ log base 3 of 3=2

$\log_{3} (3 x^{2}) + \log_{3} (3) = 2$

Simplify the left side.

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

$\log_{3} (3 x^{2} \cdot 3) = 2$

Multiply $3$ by $3$ .

$\log_{3} (9 x^{2}) = 2$

Rewrite $\log_{3} (9 x^{2}) = 2$ 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 $b^{y} = x$ .

$3^{2} = 9 x^{2}$

Solve for $x$ .

Rewrite the equation as $9 x^{2} = 3^{2}$ .

$9 x^{2} = 3^{2}$

Divide each term in $9 x^{2} = 3^{2}$ by $9$ and simplify.

Divide each term in $9 x^{2} = 3^{2}$ by $9$ .

$\frac{9 x^{2}}{9} = \frac{3^{2}}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{2}}{9} = \frac{3^{2}}{9}$

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

$x^{2} = \frac{3^{2}}{9}$

Simplify the right side.

Raise $3$ to the power of $2$ .

$x^{2} = \frac{9}{9}$

Divide $9$ by $9$ .

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

Solve for x log base 3 of 3x^2+ log base 3 of 3=2

Related Questions

Support LCMGCF.com by disabling your adblocker.