Calculus Examples

$\ln (x - 2) + \ln (2 x - 3) = 2 \ln (x)$

Step 1

Simplify the left side.

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Step 1.1

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

$\ln ((x - 2) (2 x - 3)) = 2 \ln (x)$

Step 1.2

Expand $(x - 2) (2 x - 3)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\ln (x (2 x - 3) - 2 (2 x - 3)) = 2 \ln (x)$

Step 1.2.2

Apply the distributive property.

$\ln (x (2 x) + x \cdot - 3 - 2 (2 x - 3)) = 2 \ln (x)$

Step 1.2.3

Apply the distributive property.

$\ln (x (2 x) + x \cdot - 3 - 2 (2 x) - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Rewrite using the commutative property of multiplication.

$\ln (2 x \cdot x + x \cdot - 3 - 2 (2 x) - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.3.1.2.1

Move $x$ .

$\ln (2 (x \cdot x) + x \cdot - 3 - 2 (2 x) - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3.1.2.2

Multiply $x$ by $x$ .

$\ln (2 x^{2} + x \cdot - 3 - 2 (2 x) - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3.1.3

Move $- 3$ to the left of $x$ .

$\ln (2 x^{2} - 3 \cdot x - 2 (2 x) - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3.1.4

Multiply $2$ by $- 2$ .

$\ln (2 x^{2} - 3 x - 4 x - 2 \cdot - 3) = 2 \ln (x)$

Step 1.3.1.5

Multiply $- 2$ by $- 3$ .

$\ln (2 x^{2} - 3 x - 4 x + 6) = 2 \ln (x)$

Step 1.3.2

Subtract $4 x$ from $- 3 x$ .

$\ln (2 x^{2} - 7 x + 6) = 2 \ln (x)$

Step 2

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

$\ln (2 x^{2} - 7 x + 6) = \ln (x^{2})$

Step 3

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

$2 x^{2} - 7 x + 6 = x^{2}$

Step 4

Solve for $x$ .

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Step 4.1

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

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Step 4.1.1

Subtract $x^{2}$ from both sides of the equation.

$2 x^{2} - 7 x + 6 - x^{2} = 0$

Step 4.1.2

Subtract $x^{2}$ from $2 x^{2}$ .

$x^{2} - 7 x + 6 = 0$

Step 4.2

Factor $x^{2} - 7 x + 6$ using the AC method.

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Step 4.2.1

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 $6$ and whose sum is $- 7$ .

$- 6, - 1$

Step 4.2.2

Write the factored form using these integers.

$(x - 6) (x - 1) = 0$

Step 4.3

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

$x - 6 = 0$

$x - 1 = 0$

Step 4.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 4.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 4.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 4.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.5.2

Add $1$ to both sides of the equation.

$x = 1$

Step 4.6

The final solution is all the values that make $(x - 6) (x - 1) = 0$ true.

$x = 6, 1$

Step 5

Exclude the solutions that do not make $\ln (x - 2) + \ln (2 x - 3) = 2 \ln (x)$ true.

$x = 6$