Precalculus Examples

$\log (8 x - 5) = \log (x + 1) + \log (9)$

Step 1

Simplify the right side.

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Step 1.1

Simplify $\log (x + 1) + \log (9)$ .

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Step 1.1.1

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

$\log (8 x - 5) = \log ((x + 1) \cdot 9)$

Step 1.1.2

Apply the distributive property.

$\log (8 x - 5) = \log (x \cdot 9 + 1 \cdot 9)$

Step 1.1.3

Simplify the expression.

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Step 1.1.3.1

Move $9$ to the left of $x$ .

$\log (8 x - 5) = \log (9 \cdot x + 1 \cdot 9)$

Step 1.1.3.2

Multiply $9$ by $1$ .

$\log (8 x - 5) = \log (9 x + 9)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$8 x - 5 = 9 x + 9$

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

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

$8 x - 5 - 9 x = 9$

Step 3.1.2

Subtract $9 x$ from $8 x$ .

$- x - 5 = 9$

Step 3.2

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

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Step 3.2.1

Add $5$ to both sides of the equation.

$- x = 9 + 5$

Step 3.2.2

Add $9$ and $5$ .

$- x = 14$

Step 3.3

Divide each term in $- x = 14$ by $- 1$ and simplify.

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Step 3.3.1

Divide each term in $- x = 14$ by $- 1$ .

$\frac{- x}{- 1} = \frac{14}{- 1}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{14}{- 1}$

Step 3.3.2.2

Divide $x$ by $1$ .

$x = \frac{14}{- 1}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $14$ by $- 1$ .

$x = - 14$

Step 4

Exclude the solutions that do not make $\log (8 x - 5) = \log (x + 1) + \log (9)$ true.

$No solution$