Algebra Examples

$2 \ln (x) - \ln (2 x - 7) = \ln (5 x) - \ln (x - 2)$

Step 1

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

$2 \ln (x) - \ln (2 x - 7) = \ln (\frac{5 x}{x - 2})$

Step 2

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

$2 \ln (x) - \ln (2 x - 7) - \ln (\frac{5 x}{x - 2}) = 0$

Step 3

Simplify the left side.

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Step 3.1

Simplify $2 \ln (x) - \ln (2 x - 7) - \ln (\frac{5 x}{x - 2})$ .

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Step 3.1.1

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$\ln (x^{2}) - \ln (2 x - 7) - \ln (\frac{5 x}{x - 2}) = 0$

Step 3.1.2

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

$\ln (\frac{x^{2}}{2 x - 7}) - \ln (\frac{5 x}{x - 2}) = 0$

Step 3.1.3

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

$\ln (\frac{\frac{x^{2}}{2 x - 7}}{\frac{5 x}{x - 2}}) = 0$

Step 3.1.4

Multiply the numerator by the reciprocal of the denominator.

$\ln (\frac{x^{2}}{2 x - 7} \cdot \frac{x - 2}{5 x}) = 0$

Step 3.1.5

Cancel the common factor of $x$ .

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Step 3.1.5.1

Factor $x$ out of $x^{2}$ .

$\ln (\frac{x \cdot x}{2 x - 7} \cdot \frac{x - 2}{5 x}) = 0$

Step 3.1.5.2

Factor $x$ out of $5 x$ .

$\ln (\frac{x \cdot x}{2 x - 7} \cdot \frac{x - 2}{x \cdot 5}) = 0$

Step 3.1.5.3

Cancel the common factor.

$\ln (\frac{x \cdot x}{2 x - 7} \cdot \frac{x - 2}{x \cdot 5}) = 0$

Step 3.1.5.4

Rewrite the expression.

$\ln (\frac{x}{2 x - 7} \cdot \frac{x - 2}{5}) = 0$

Step 3.1.6

Multiply $\frac{x}{2 x - 7}$ by $\frac{x - 2}{5}$ .

$\ln (\frac{x (x - 2)}{(2 x - 7) \cdot 5}) = 0$

Step 3.1.7

Move $5$ to the left of $2 x - 7$ .

$\ln (\frac{x (x - 2)}{5 (2 x - 7)}) = 0$

Step 4

Rewrite $\ln (\frac{x (x - 2)}{5 (2 x - 7)}) = 0$ 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 $b^{y} = x$ .

$e^{0} = \frac{x (x - 2)}{5 (2 x - 7)}$

Step 5

Cross multiply to remove the fraction.

$x (x - 2) = e^{0} (5 (2 x - 7))$

Step 6

Simplify $e^{0} (5 (2 x - 7))$ .

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Step 6.1

Simplify the expression.

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Step 6.1.1

Anything raised to $0$ is $1$ .

$x (x - 2) = 1 (5 (2 x - 7))$

Step 6.1.2

Multiply $5 (2 x - 7)$ by $1$ .

$x (x - 2) = 5 (2 x - 7)$

Step 6.2

Apply the distributive property.

$x (x - 2) = 5 (2 x) + 5 \cdot - 7$

Step 6.3

Multiply.

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Step 6.3.1

Multiply $2$ by $5$ .

$x (x - 2) = 10 x + 5 \cdot - 7$

Step 6.3.2

Multiply $5$ by $- 7$ .

$x (x - 2) = 10 x - 35$

Step 7

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

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Step 7.1

Subtract $10 x$ from both sides of the equation.

$x (x - 2) - 10 x = - 35$

Step 7.2

Simplify each term.

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Step 7.2.1

Apply the distributive property.

$x \cdot x + x \cdot - 2 - 10 x = - 35$

Step 7.2.2

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 2 - 10 x = - 35$

Step 7.2.3

Move $- 2$ to the left of $x$ .

$x^{2} - 2 x - 10 x = - 35$

Step 7.3

Subtract $10 x$ from $- 2 x$ .

$x^{2} - 12 x = - 35$

Step 8

Factor $x$ out of $x^{2} - 12 x$ .

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Step 8.1

Factor $x$ out of $x^{2}$ .

$x \cdot x - 12 x = - 35$

Step 8.2

Factor $x$ out of $- 12 x$ .

$x \cdot x + x \cdot - 12 = - 35$

Step 8.3

Factor $x$ out of $x \cdot x + x \cdot - 12$ .

$x (x - 12) = - 35$

Step 9

Simplify $x (x - 12)$ .

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Step 9.1

Apply the distributive property.

$x \cdot x + x \cdot - 12 = - 35$

Step 9.2

Simplify the expression.

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Step 9.2.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 12 = - 35$

Step 9.2.2

Move $- 12$ to the left of $x$ .

$x^{2} - 12 x = - 35$

Step 10

Add $35$ to both sides of the equation.

$x^{2} - 12 x + 35 = 0$

Step 11

Factor $x^{2} - 12 x + 35$ using the AC method.

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Step 11.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $35$ and whose sum is $- 12$ .

$- 7, - 5$

Step 11.2

Write the factored form using these integers.

$(x - 7) (x - 5) = 0$

Step 12

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 7 = 0$

$x - 5 = 0$

Step 13

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 13.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 13.2

Add $7$ to both sides of the equation.

$x = 7$

Step 14

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 14.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 14.2

Add $5$ to both sides of the equation.

$x = 5$

Step 15

The final solution is all the values that make $(x - 7) (x - 5) = 0$ true.

$x = 7, 5$