Precalculus Examples

$\log (2 x - 1) + \log (x + 4) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((2 x - 1) (x + 4)) = 1$

Step 1.2

Expand $(2 x - 1) (x + 4)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\log (2 x (x + 4) - 1 (x + 4)) = 1$

Step 1.2.2

Apply the distributive property.

$\log (2 x \cdot x + 2 x \cdot 4 - 1 (x + 4)) = 1$

Step 1.2.3

Apply the distributive property.

$\log (2 x \cdot x + 2 x \cdot 4 - 1 x - 1 \cdot 4) = 1$

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 $x$ .

$\log (2 (x \cdot x) + 2 x \cdot 4 - 1 x - 1 \cdot 4) = 1$

Step 1.3.1.1.2

Multiply $x$ by $x$ .

$\log (2 x^{2} + 2 x \cdot 4 - 1 x - 1 \cdot 4) = 1$

Step 1.3.1.2

Multiply $4$ by $2$ .

$\log (2 x^{2} + 8 x - 1 x - 1 \cdot 4) = 1$

Step 1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\log (2 x^{2} + 8 x - x - 1 \cdot 4) = 1$

Step 1.3.1.4

Multiply $- 1$ by $4$ .

$\log (2 x^{2} + 8 x - x - 4) = 1$

Step 1.3.2

Subtract $x$ from $8 x$ .

$\log (2 x^{2} + 7 x - 4) = 1$

Step 2

Rewrite $\log (2 x^{2} + 7 x - 4) = 1$ 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 $b^{y} = x$ .

$10^{1} = 2 x^{2} + 7 x - 4$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $2 x^{2} + 7 x - 4 = 10^{1}$ .

$2 x^{2} + 7 x - 4 = 10$

Step 3.2

Move all terms to the left side of the equation and simplify.

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Step 3.2.1

Subtract $10$ from both sides of the equation.

$2 x^{2} + 7 x - 4 - 10 = 0$

Step 3.2.2

Subtract $10$ from $- 4$ .

$2 x^{2} + 7 x - 14 = 0$

Step 3.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.4

Substitute the values $a = 2$ , $b = 7$ , and $c = - 14$ into the quadratic formula and solve for $x$ .

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (2 \cdot - 14)}}{2 \cdot 2}$

Step 3.5

Simplify.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Raise $7$ to the power of $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 2 \cdot - 14}}{2 \cdot 2}$

Step 3.5.1.2

Multiply $- 4 \cdot 2 \cdot - 14$ .

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Step 3.5.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 7 \pm \sqrt{49 - 8 \cdot - 14}}{2 \cdot 2}$

Step 3.5.1.2.2

Multiply $- 8$ by $- 14$ .

$x = \frac{- 7 \pm \sqrt{49 + 112}}{2 \cdot 2}$

Step 3.5.1.3

Add $49$ and $112$ .

$x = \frac{- 7 \pm \sqrt{161}}{2 \cdot 2}$

Step 3.5.2

Multiply $2$ by $2$ .

$x = \frac{- 7 \pm \sqrt{161}}{4}$

Step 3.6

The final answer is the combination of both solutions.

$x = - \frac{7 - \sqrt{161}}{4}, - \frac{7 + \sqrt{161}}{4}$

Step 4

Exclude the solutions that do not make $\log (2 x - 1) + \log (x + 4) = 1$ true.

$x = - \frac{7 - \sqrt{161}}{4}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{7 - \sqrt{161}}{4}$

Decimal Form:

$x = 1.42214438 \dots$