Calculus Examples

$\log (w) + \log (3 w - 13) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log (w (3 w - 13)) = 1$

Step 1.2

Simplify by multiplying through.

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Step 1.2.1

Apply the distributive property.

$\log (w (3 w) + w \cdot - 13) = 1$

Step 1.2.2

Reorder.

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Step 1.2.2.1

Rewrite using the commutative property of multiplication.

$\log (3 w \cdot w + w \cdot - 13) = 1$

Step 1.2.2.2

Move $- 13$ to the left of $w$ .

$\log (3 w \cdot w - 13 \cdot w) = 1$

Step 1.3

Multiply $w$ by $w$ by adding the exponents.

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Step 1.3.1

Move $w$ .

$\log (3 (w \cdot w) - 13 \cdot w) = 1$

Step 1.3.2

Multiply $w$ by $w$ .

$\log (3 w^{2} - 13 \cdot w) = 1$

$\log (3 w^{2} - 13 w) = 1$

Step 2

Rewrite $\log (3 w^{2} - 13 w) = 1$ 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 $b^{y} = x$ .

$10^{1} = 3 w^{2} - 13 w$

Step 3

Solve for $w$ .

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Step 3.1

Rewrite the equation as $3 w^{2} - 13 w = 10^{1}$ .

$3 w^{2} - 13 w = 10$

Step 3.2

Subtract $10$ from both sides of the equation.

$3 w^{2} - 13 w - 10 = 0$

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 $a \cdot c = 3 \cdot - 10 = - 30$ and whose sum is $b = - 13$ .

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Step 3.3.1.1

Factor $- 13$ out of $- 13 w$ .

$3 w^{2} - 13 w - 10 = 0$

Step 3.3.1.2

Rewrite $- 13$ as $2$ plus $- 15$

$3 w^{2} + (2 - 15) w - 10 = 0$

Step 3.3.1.3

Apply the distributive property.

$3 w^{2} + 2 w - 15 w - 10 = 0$

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.

$(3 w^{2} + 2 w) - 15 w - 10 = 0$

Step 3.3.2.2

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

$w (3 w + 2) - 5 (3 w + 2) = 0$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $3 w + 2$ .

$(3 w + 2) (w - 5) = 0$

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 $0$ .

$3 w + 2 = 0$

$w - 5 = 0$

Step 3.5

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

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Step 3.5.1

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

$3 w + 2 = 0$

Step 3.5.2

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

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Step 3.5.2.1

Subtract $2$ from both sides of the equation.

$3 w = - 2$

Step 3.5.2.2

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

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Step 3.5.2.2.1

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

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

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

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

Step 3.5.2.2.2.1.2

Divide $w$ by $1$ .

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

Step 3.5.2.2.3

Simplify the right side.

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Step 3.5.2.2.3.1

Move the negative in front of the fraction.

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

Step 3.6

Set $w - 5$ equal to $0$ and solve for $w$ .

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Step 3.6.1

Set $w - 5$ equal to $0$ .

$w - 5 = 0$

Step 3.6.2

Add $5$ to both sides of the equation.

$w = 5$

Step 3.7

The final solution is all the values that make $(3 w + 2) (w - 5) = 0$ true.

$w = - \frac{2}{3}, 5$

Step 4

Exclude the solutions that do not make $\log (w) + \log (3 w - 13) = 1$ true.

$w = 5$