Calculus Examples

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

Step 1

Simplify the left side.

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Step 1.1

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

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

Step 1.2

Apply the distributive property.

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

Step 1.3

Simplify the expression.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

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

Step 1.3.2

Multiply $x$ by $1$ .

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

Step 1.4

Multiply $x$ by $x$ by adding the exponents.

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Step 1.4.1

Move $x$ .

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

Step 1.4.2

Multiply $x$ by $x$ .

$\log_{3} (2 x^{2} + x) = 1$

Step 2

Rewrite $\log_{3} (2 x^{2} + x) = 1$ 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 $b^{y} = x$ .

$3^{1} = 2 x^{2} + x$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $2 x^{2} + x = 3^{1}$ .

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

Step 3.2

Subtract $3$ from both sides of the equation.

$2 x^{2} + x - 3 = 0$

Step 3.3

Factor by grouping.

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Step 3.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 1$ .

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Step 3.3.1.1

Multiply by $1$ .

$2 x^{2} + 1 x - 3 = 0$

Step 3.3.1.2

Rewrite $1$ as $- 2$ plus $3$

$2 x^{2} + (- 2 + 3) x - 3 = 0$

Step 3.3.1.3

Apply the distributive property.

$2 x^{2} - 2 x + 3 x - 3 = 0$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 2 x) + 3 x - 3 = 0$

Step 3.3.2.2

Factor out the greatest common factor (GCF) from each group.

$2 x (x - 1) + 3 (x - 1) = 0$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (2 x + 3) = 0$

Step 3.4

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

$x - 1 = 0$

$2 x + 3 = 0$

Step 3.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 3.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.6

Set $2 x + 3$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $2 x + 3$ equal to $0$ .

$2 x + 3 = 0$

Step 3.6.2

Solve $2 x + 3 = 0$ for $x$ .

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Step 3.6.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 3.6.2.2

Divide each term in $2 x = - 3$ by $2$ and simplify.

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Step 3.6.2.2.1

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Step 3.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Step 3.6.2.2.3

Simplify the right side.

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Step 3.6.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 3.7

The final solution is all the values that make $(x - 1) (2 x + 3) = 0$ true.

$x = 1, - \frac{3}{2}$

Step 4

Exclude the solutions that do not make $\log_{3} (x) + \log_{3} (2 x + 1) = 1$ true.

$x = 1$