Calculus Examples

$\log (x + 3) + \log (4) = 2 \log (x)$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((x + 3) \cdot 4) = 2 \log (x)$

Step 1.2

Apply the distributive property.

$\log (x \cdot 4 + 3 \cdot 4) = 2 \log (x)$

Step 1.3

Simplify the expression.

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Step 1.3.1

Move $4$ to the left of $x$ .

$\log (4 \cdot x + 3 \cdot 4) = 2 \log (x)$

Step 1.3.2

Multiply $3$ by $4$ .

$\log (4 x + 12) = 2 \log (x)$

Step 2

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

$\log (4 x + 12) = \log (x^{2})$

Step 3

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

$4 x + 12 = x^{2}$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $x^{2}$ from both sides of the equation.

$4 x + 12 - x^{2} = 0$

Step 4.2

Factor the left side of the equation.

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Step 4.2.1

Factor $- 1$ out of $4 x + 12 - x^{2}$ .

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Step 4.2.1.1

Reorder the expression.

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Step 4.2.1.1.1

Move $12$ .

$4 x - x^{2} + 12 = 0$

Step 4.2.1.1.2

Reorder $4 x$ and $- x^{2}$ .

$- x^{2} + 4 x + 12 = 0$

Step 4.2.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 4 x + 12 = 0$

Step 4.2.1.3

Factor $- 1$ out of $4 x$ .

$- (x^{2}) - (- 4 x) + 12 = 0$

Step 4.2.1.4

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - (- 4 x) - 1 \cdot - 12 = 0$

Step 4.2.1.5

Factor $- 1$ out of $- (x^{2}) - (- 4 x)$ .

$- (x^{2} - 4 x) - 1 \cdot - 12 = 0$

Step 4.2.1.6

Factor $- 1$ out of $- (x^{2} - 4 x) - 1 (- 12)$ .

$- (x^{2} - 4 x - 12) = 0$

Step 4.2.2

Factor.

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Step 4.2.2.1

Factor $x^{2} - 4 x - 12$ using the AC method.

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Step 4.2.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 4.2.2.1.2

Write the factored form using these integers.

$- ((x - 6) (x + 2)) = 0$

Step 4.2.2.2

Remove unnecessary parentheses.

$- (x - 6) (x + 2) = 0$

Step 4.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x + 2 = 0$

Step 4.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.6

The final solution is all the values that make $- (x - 6) (x + 2) = 0$ true.

$x = 6, - 2$

Step 5

Exclude the solutions that do not make $\log (x + 3) + \log (4) = 2 \log (x)$ true.

$x = 6$