Precalculus Examples

$\log (3 x + 7) + \log (x - 2) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((3 x + 7) (x - 2)) = 1$

Step 1.2

Expand $(3 x + 7) (x - 2)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log (3 x (x - 2) + 7 (x - 2)) = 1$

Step 1.2.2

Apply the distributive property.

$\log (3 x \cdot x + 3 x \cdot - 2 + 7 (x - 2)) = 1$

Step 1.2.3

Apply the distributive property.

$\log (3 x \cdot x + 3 x \cdot - 2 + 7 x + 7 \cdot - 2) = 1$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

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

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Step 1.3.1.1.1

Move $x$ .

$\log (3 (x \cdot x) + 3 x \cdot - 2 + 7 x + 7 \cdot - 2) = 1$

Step 1.3.1.1.2

Multiply $x$ by $x$ .

$\log (3 x^{2} + 3 x \cdot - 2 + 7 x + 7 \cdot - 2) = 1$

Step 1.3.1.2

Multiply $- 2$ by $3$ .

$\log (3 x^{2} - 6 x + 7 x + 7 \cdot - 2) = 1$

Step 1.3.1.3

Multiply $7$ by $- 2$ .

$\log (3 x^{2} - 6 x + 7 x - 14) = 1$

Step 1.3.2

Add $- 6 x$ and $7 x$ .

$\log (3 x^{2} + x - 14) = 1$

Step 2

Rewrite $\log (3 x^{2} + x - 14) = 1$ 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 $b^{y} = x$ .

$10^{1} = 3 x^{2} + x - 14$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $3 x^{2} + x - 14 = 10^{1}$ .

$3 x^{2} + x - 14 = 10$

Step 3.2

Subtract $10$ from both sides of the equation.

$3 x^{2} + x - 14 - 10 = 0$

Step 3.3

Subtract $10$ from $- 14$ .

$3 x^{2} + x - 24 = 0$

Step 3.4

Factor by grouping.

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Step 3.4.1

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 $a \cdot c = 3 \cdot - 24 = - 72$ and whose sum is $b = 1$ .

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Step 3.4.1.1

Multiply by $1$ .

$3 x^{2} + 1 x - 24 = 0$

Step 3.4.1.2

Rewrite $1$ as $- 8$ plus $9$

$3 x^{2} + (- 8 + 9) x - 24 = 0$

Step 3.4.1.3

Apply the distributive property.

$3 x^{2} - 8 x + 9 x - 24 = 0$

Step 3.4.2

Factor out the greatest common factor from each group.

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Step 3.4.2.1

Group the first two terms and the last two terms.

$(3 x^{2} - 8 x) + 9 x - 24 = 0$

Step 3.4.2.2

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

$x (3 x - 8) + 3 (3 x - 8) = 0$

Step 3.4.3

Factor the polynomial by factoring out the greatest common factor, $3 x - 8$ .

$(3 x - 8) (x + 3) = 0$

Step 3.5

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

$3 x - 8 = 0$

$x + 3 = 0$

Step 3.6

Set $3 x - 8$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $3 x - 8$ equal to $0$ .

$3 x - 8 = 0$

Step 3.6.2

Solve $3 x - 8 = 0$ for $x$ .

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Step 3.6.2.1

Add $8$ to both sides of the equation.

$3 x = 8$

Step 3.6.2.2

Divide each term in $3 x = 8$ by $3$ and simplify.

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Step 3.6.2.2.1

Divide each term in $3 x = 8$ by $3$ .

$\frac{3 x}{3} = \frac{8}{3}$

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 $3$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{8}{3}$

Step 3.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{8}{3}$

Step 3.7

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

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Step 3.7.1

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

$x + 3 = 0$

Step 3.7.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.8

The final solution is all the values that make $(3 x - 8) (x + 3) = 0$ true.

$x = \frac{8}{3}, - 3$

Step 4

Exclude the solutions that do not make $\log (3 x + 7) + \log (x - 2) = 1$ true.

$x = \frac{8}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{8}{3}$

Decimal Form:

$x = 2.\overline{6}$

Mixed Number Form:

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