Precalculus Examples

$\log_{2} (y - 7) = 3 + \log_{2} (3 y + 1)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{2} (y - 7) - \log_{2} (3 y + 1) = 3$

Step 2

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

$\log_{2} (\frac{y - 7}{3 y + 1}) = 3$

Step 3

Rewrite $\log_{2} (\frac{y - 7}{3 y + 1}) = 3$ 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 $b^{y} = x$ .

$2^{3} = \frac{y - 7}{3 y + 1}$

Step 4

Cross multiply to remove the fraction.

$y - 7 = 2^{3} (3 y + 1)$

Step 5

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

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Step 5.1

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

$y - 7 = 8 (3 y + 1)$

Step 5.2

Apply the distributive property.

$y - 7 = 8 (3 y) + 8 \cdot 1$

Step 5.3

Multiply.

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Step 5.3.1

Multiply $3$ by $8$ .

$y - 7 = 24 y + 8 \cdot 1$

Step 5.3.2

Multiply $8$ by $1$ .

$y - 7 = 24 y + 8$

Step 6

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

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Step 6.1

Subtract $24 y$ from both sides of the equation.

$y - 7 - 24 y = 8$

Step 6.2

Subtract $24 y$ from $y$ .

$- 23 y - 7 = 8$

Step 7

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

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Step 7.1

Add $7$ to both sides of the equation.

$- 23 y = 8 + 7$

Step 7.2

Add $8$ and $7$ .

$- 23 y = 15$

Step 8

Divide each term in $- 23 y = 15$ by $- 23$ and simplify.

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Step 8.1

Divide each term in $- 23 y = 15$ by $- 23$ .

$\frac{- 23 y}{- 23} = \frac{15}{- 23}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $- 23$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{- 23 y}{- 23} = \frac{15}{- 23}$

Step 8.2.1.2

Divide $y$ by $1$ .

$y = \frac{15}{- 23}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Move the negative in front of the fraction.

$y = - \frac{15}{23}$

Step 9

Exclude the solutions that do not make $\log_{2} (y - 7) = 3 + \log_{2} (3 y + 1)$ true.

$No solution$