Precalculus Examples

$2 \log_{5} (x) = \log_{5} (2) + \log_{5} (4 + 10)$

Step 1

Simplify the left side.

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Step 1.1

Simplify $2 \log_{5} (x)$ by moving $2$ inside the logarithm.

$\log_{5} (x^{2}) = \log_{5} (2) + \log_{5} (4 + 10)$

Step 2

Simplify the right side.

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Step 2.1

Simplify $\log_{5} (2) + \log_{5} (4 + 10)$ .

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Step 2.1.1

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

$\log_{5} (x^{2}) = \log_{5} (2 (4 + 10))$

Step 2.1.2

Add $4$ and $10$ .

$\log_{5} (x^{2}) = \log_{5} (2 \cdot 14)$

Step 2.1.3

Multiply $2$ by $14$ .

$\log_{5} (x^{2}) = \log_{5} (28)$

Step 3

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

$x^{2} = 28$

Step 4

Solve for $x$ .

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Step 4.1

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

$x = \pm \sqrt{28}$

Step 4.2

Simplify $\pm \sqrt{28}$ .

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Step 4.2.1

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 4.2.1.1

Factor $4$ out of $28$ .

$x = \pm \sqrt{4 (7)}$

Step 4.2.1.2

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

$x = \pm \sqrt{2^{2} \cdot 7}$

Step 4.2.2

Pull terms out from under the radical.

$x = \pm 2 \sqrt{7}$

Step 4.3

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

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Step 4.3.1

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

$x = 2 \sqrt{7}$

Step 4.3.2

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

$x = - 2 \sqrt{7}$

Step 4.3.3

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

$x = 2 \sqrt{7}, - 2 \sqrt{7}$

Step 5

Exclude the solutions that do not make $2 \log_{5} (x) = \log_{5} (2) + \log_{5} (4 + 10)$ true.

$x = 2 \sqrt{7}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = 2 \sqrt{7}$

Decimal Form:

$x = 5.29150262 \dots$