Precalculus Examples

$\log (2 x - 1) + \log (5) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log ((2 x - 1) \cdot 5) = 1$

Step 1.2

Apply the distributive property.

$\log (2 x \cdot 5 - 1 \cdot 5) = 1$

Step 1.3

Multiply.

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Step 1.3.1

Multiply $5$ by $2$ .

$\log (10 x - 1 \cdot 5) = 1$

Step 1.3.2

Multiply $- 1$ by $5$ .

$\log (10 x - 5) = 1$

Step 2

Rewrite $\log (10 x - 5) = 1$ 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 $b^{y} = x$ .

$10^{1} = 10 x - 5$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $10 x - 5 = 10^{1}$ .

$10 x - 5 = 10$

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.

$10 x = 10 + 5$

Step 3.2.2

Add $10$ and $5$ .

$10 x = 15$

Step 3.3

Divide each term in $10 x = 15$ by $10$ and simplify.

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Step 3.3.1

Divide each term in $10 x = 15$ by $10$ .

$\frac{10 x}{10} = \frac{15}{10}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $10$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{10 x}{10} = \frac{15}{10}$

Step 3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{15}{10}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Cancel the common factor of $15$ and $10$ .

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Step 3.3.3.1.1

Factor $5$ out of $15$ .

$x = \frac{5 (3)}{10}$

Step 3.3.3.1.2

Cancel the common factors.

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Step 3.3.3.1.2.1

Factor $5$ out of $10$ .

$x = \frac{5 \cdot 3}{5 \cdot 2}$

Step 3.3.3.1.2.2

Cancel the common factor.

$x = \frac{5 \cdot 3}{5 \cdot 2}$

Step 3.3.3.1.2.3

Rewrite the expression.

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.5$

Mixed Number Form:

$x = 1 \frac{1}{2}$