Precalculus Examples

$\ln (3 y + 5) - \ln (y - 1) = \ln (- 4 y + 10)$

Step 1

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

$\ln (\frac{3 y + 5}{y - 1}) = \ln (- 4 y + 10)$

Step 2

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

$\frac{3 y + 5}{y - 1} = - 4 y + 10$

Step 3

Solve for $y$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y - 1, 1, 1$

Step 3.1.2

Remove parentheses.

$y - 1, 1, 1$

Step 3.1.3

The LCM of one and any expression is the expression.

$y - 1$

Step 3.2

Multiply each term in $\frac{3 y + 5}{y - 1} = - 4 y + 10$ by $y - 1$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{3 y + 5}{y - 1} = - 4 y + 10$ by $y - 1$ .

$\frac{3 y + 5}{y - 1} (y - 1) = - 4 y (y - 1) + 10 (y - 1)$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $y - 1$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 y + 5}{y - 1} (y - 1) = - 4 y (y - 1) + 10 (y - 1)$

Step 3.2.2.1.2

Rewrite the expression.

$3 y + 5 = - 4 y (y - 1) + 10 (y - 1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Simplify each term.

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Step 3.2.3.1.1

Apply the distributive property.

$3 y + 5 = - 4 y \cdot y - 4 y \cdot - 1 + 10 (y - 1)$

Step 3.2.3.1.2

Multiply $y$ by $y$ by adding the exponents.

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Step 3.2.3.1.2.1

Move $y$ .

$3 y + 5 = - 4 (y \cdot y) - 4 y \cdot - 1 + 10 (y - 1)$

Step 3.2.3.1.2.2

Multiply $y$ by $y$ .

$3 y + 5 = - 4 y^{2} - 4 y \cdot - 1 + 10 (y - 1)$

Step 3.2.3.1.3

Multiply $- 1$ by $- 4$ .

$3 y + 5 = - 4 y^{2} + 4 y + 10 (y - 1)$

Step 3.2.3.1.4

Apply the distributive property.

$3 y + 5 = - 4 y^{2} + 4 y + 10 y + 10 \cdot - 1$

Step 3.2.3.1.5

Multiply $10$ by $- 1$ .

$3 y + 5 = - 4 y^{2} + 4 y + 10 y - 10$

Step 3.2.3.2

Add $4 y$ and $10 y$ .

$3 y + 5 = - 4 y^{2} + 14 y - 10$

Step 3.3

Solve the equation.

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Step 3.3.1

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 4 y^{2} + 14 y - 10 = 3 y + 5$

Step 3.3.2

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

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Step 3.3.2.1

Subtract $3 y$ from both sides of the equation.

$- 4 y^{2} + 14 y - 10 - 3 y = 5$

Step 3.3.2.2

Subtract $3 y$ from $14 y$ .

$- 4 y^{2} + 11 y - 10 = 5$

Step 3.3.3

Move all terms to the left side of the equation and simplify.

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Step 3.3.3.1

Subtract $5$ from both sides of the equation.

$- 4 y^{2} + 11 y - 10 - 5 = 0$

Step 3.3.3.2

Subtract $5$ from $- 10$ .

$- 4 y^{2} + 11 y - 15 = 0$

Step 3.3.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.3.5

Substitute the values $a = - 4$ , $b = 11$ , and $c = - 15$ into the quadratic formula and solve for $y$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (- 4 \cdot - 15)}}{2 \cdot - 4}$

Step 3.3.6

Simplify.

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Step 3.3.6.1

Simplify the numerator.

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Step 3.3.6.1.1

Raise $11$ to the power of $2$ .

$y = \frac{- 11 \pm \sqrt{121 - 4 \cdot - 4 \cdot - 15}}{2 \cdot - 4}$

Step 3.3.6.1.2

Multiply $- 4 \cdot - 4 \cdot - 15$ .

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Step 3.3.6.1.2.1

Multiply $- 4$ by $- 4$ .

$y = \frac{- 11 \pm \sqrt{121 + 16 \cdot - 15}}{2 \cdot - 4}$

Step 3.3.6.1.2.2

Multiply $16$ by $- 15$ .

$y = \frac{- 11 \pm \sqrt{121 - 240}}{2 \cdot - 4}$

Step 3.3.6.1.3

Subtract $240$ from $121$ .

$y = \frac{- 11 \pm \sqrt{- 119}}{2 \cdot - 4}$

Step 3.3.6.1.4

Rewrite $- 119$ as $- 1 (119)$ .

$y = \frac{- 11 \pm \sqrt{- 1 \cdot 119}}{2 \cdot - 4}$

Step 3.3.6.1.5

Rewrite $\sqrt{- 1 (119)}$ as $\sqrt{- 1} \cdot \sqrt{119}$ .

$y = \frac{- 11 \pm \sqrt{- 1} \cdot \sqrt{119}}{2 \cdot - 4}$

Step 3.3.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{- 11 \pm i \sqrt{119}}{2 \cdot - 4}$

Step 3.3.6.2

Multiply $2$ by $- 4$ .

$y = \frac{- 11 \pm i \sqrt{119}}{- 8}$

Step 3.3.6.3

Simplify $\frac{- 11 \pm i \sqrt{119}}{- 8}$ .

$y = \frac{11 \pm i \sqrt{119}}{8}$

Step 3.3.7

The final answer is the combination of both solutions.

$y = \frac{11 + i \sqrt{119}}{8}, \frac{11 - i \sqrt{119}}{8}$