Precalculus Examples

$2 \log (3) + \frac{1}{2} \cdot \log (16) - \log (3) = \log (x)$

Step 1

Rewrite the equation as $\log (x) = 2 \log (3) + \frac{1}{2} \cdot \log (16) - \log (3)$ .

$\log (x) = 2 \log (3) + \frac{1}{2} \cdot \log (16) - \log (3)$

Step 2

Simplify the right side.

Tap for more steps...

Step 2.1

Simplify $2 \log (3) + \frac{1}{2} \log (16) - \log (3)$ .

Tap for more steps...

Step 2.1.1

Combine $\frac{1}{2}$ and $\log (16)$ .

$\log (x) = 2 \log (3) + \frac{\log (16)}{2} - \log (3)$

Step 2.1.2

Subtract $\log (3)$ from $2 \log (3)$ .

$\log (x) = \log (3) + \frac{\log (16)}{2}$

Step 3

Simplify the right side.

Tap for more steps...

Step 3.1

Simplify $\log (3) + \frac{\log (16)}{2}$ .

Tap for more steps...

Step 3.1.1

Simplify each term.

Tap for more steps...

Step 3.1.1.1

Rewrite $\log (16)$ as $\log (2^{4})$ .

$\log (x) = \log (3) + \frac{\log (2^{4})}{2}$

Step 3.1.1.2

Expand $\log (2^{4})$ by moving $4$ outside the logarithm.

$\log (x) = \log (3) + \frac{4 \log (2)}{2}$

Step 3.1.1.3

Cancel the common factor of $4$ and $2$ .

Tap for more steps...

Step 3.1.1.3.1

Factor $2$ out of $4 \log (2)$ .

$\log (x) = \log (3) + \frac{2 (2 \log (2))}{2}$

Step 3.1.1.3.2

Cancel the common factors.

Tap for more steps...

Step 3.1.1.3.2.1

Factor $2$ out of $2$ .

$\log (x) = \log (3) + \frac{2 (2 \log (2))}{2 (1)}$

Step 3.1.1.3.2.2

Cancel the common factor.

$\log (x) = \log (3) + \frac{2 (2 \log (2))}{2 \cdot 1}$

Step 3.1.1.3.2.3

Rewrite the expression.

$\log (x) = \log (3) + \frac{2 \log (2)}{1}$

Step 3.1.1.3.2.4

Divide $2 \log (2)$ by $1$ .

$\log (x) = \log (3) + 2 \log (2)$

Step 3.1.1.4

Simplify $2 \log (2)$ by moving $2$ inside the logarithm.

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

Step 3.1.1.5

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

$\log (x) = \log (3) + \log (4)$

Step 3.1.2

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

$\log (x) = \log (3 \cdot 4)$

Step 3.1.3

Multiply $3$ by $4$ .

$\log (x) = \log (12)$

Step 4

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

$x = 12$