Precalculus Examples

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

Step 1

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

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

Step 2

Logarithm base $2$ of $2$ is $1$ .

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

Step 3

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

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

Step 4

Cross multiply to remove the fraction.

$5 x + 1 = 2 (2 x + 3)$

Step 5

Simplify $2 (2 x + 3)$ .

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Step 5.1

Apply the distributive property.

$5 x + 1 = 2 (2 x) + 2 \cdot 3$

Step 5.2

Multiply.

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Step 5.2.1

Multiply $2$ by $2$ .

$5 x + 1 = 4 x + 2 \cdot 3$

Step 5.2.2

Multiply $2$ by $3$ .

$5 x + 1 = 4 x + 6$

Step 6

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

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Step 6.1

Subtract $4 x$ from both sides of the equation.

$5 x + 1 - 4 x = 6$

Step 6.2

Subtract $4 x$ from $5 x$ .

$x + 1 = 6$

Step 7

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

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Step 7.1

Subtract $1$ from both sides of the equation.

$x = 6 - 1$

Step 7.2

Subtract $1$ from $6$ .

$x = 5$