Precalculus Examples

$\log_{3} (2 x - 5) - \log_{3} (x + 1) = 2$

Step 1

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{3} (\frac{2 x - 5}{x + 1}) = 2$

Step 2

Rewrite $\log_{3} (\frac{2 x - 5}{x + 1}) = 2$ 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 $b^{y} = x$ .

$3^{2} = \frac{2 x - 5}{x + 1}$

Step 3

Cross multiply to remove the fraction.

$2 x - 5 = 3^{2} (x + 1)$

Step 4

Simplify $3^{2} (x + 1)$ .

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Step 4.1

Raise $3$ to the power of $2$ .

$2 x - 5 = 9 (x + 1)$

Step 4.2

Apply the distributive property.

$2 x - 5 = 9 x + 9 \cdot 1$

Step 4.3

Multiply $9$ by $1$ .

$2 x - 5 = 9 x + 9$

Step 5

Move all terms containing $x$ to the left side of the equation.

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Step 5.1

Subtract $9 x$ from both sides of the equation.

$2 x - 5 - 9 x = 9$

Step 5.2

Subtract $9 x$ from $2 x$ .

$- 7 x - 5 = 9$

Step 6

Move all terms not containing $x$ to the right side of the equation.

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Step 6.1

Add $5$ to both sides of the equation.

$- 7 x = 9 + 5$

Step 6.2

Add $9$ and $5$ .

$- 7 x = 14$

Step 7

Divide each term in $- 7 x = 14$ by $- 7$ and simplify.

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Step 7.1

Divide each term in $- 7 x = 14$ by $- 7$ .

$\frac{- 7 x}{- 7} = \frac{14}{- 7}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $- 7$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{- 7 x}{- 7} = \frac{14}{- 7}$

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{14}{- 7}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $14$ by $- 7$ .

$x = - 2$

Step 8

Exclude the solutions that do not make $\log_{3} (2 x - 5) - \log_{3} (x + 1) = 2$ true.

$No solution$