Precalculus Examples

$\ln (x - 5) + \ln (4) = \ln (x - \ln (2))$

Step 1

Simplify the left side.

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Step 1.1

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

$\ln ((x - 5) \cdot 4) = \ln (x - \ln (2))$

Step 1.2

Apply the distributive property.

$\ln (x \cdot 4 - 5 \cdot 4) = \ln (x - \ln (2))$

Step 1.3

Simplify the expression.

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Step 1.3.1

Move $4$ to the left of $x$ .

$\ln (4 \cdot x - 5 \cdot 4) = \ln (x - \ln (2))$

Step 1.3.2

Multiply $- 5$ by $4$ .

$\ln (4 x - 20) = \ln (x - \ln (2))$

Step 2

Simplify $- \ln (2)$ by moving $- 1$ inside the logarithm.

$\ln (4 x - 20) = \ln (x + \ln (2^{- 1}))$

Step 3

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

$4 x - 20 = x + \ln (2^{- 1})$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$4 x - 20 = x + \ln (\frac{1}{2})$

Step 4.2

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

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Step 4.2.1

Subtract $x$ from both sides of the equation.

$4 x - 20 - x = \ln (\frac{1}{2})$

Step 4.2.2

Subtract $x$ from $4 x$ .

$3 x - 20 = \ln (\frac{1}{2})$

Step 4.3

Add $20$ to both sides of the equation.

$3 x = \ln (\frac{1}{2}) + 20$

Step 4.4

Divide each term in $3 x = \ln (\frac{1}{2}) + 20$ by $3$ and simplify.

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Step 4.4.1

Divide each term in $3 x = \ln (\frac{1}{2}) + 20$ by $3$ .

$\frac{3 x}{3} = \frac{\ln (\frac{1}{2})}{3} + \frac{20}{3}$

Step 4.4.2

Simplify the left side.

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Step 4.4.2.1

Cancel the common factor of $3$ .

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Step 4.4.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\ln (\frac{1}{2})}{3} + \frac{20}{3}$

Step 4.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\ln (\frac{1}{2})}{3} + \frac{20}{3}$

Step 4.4.3

Simplify the right side.

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Step 4.4.3.1

Simplify each term.

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Step 4.4.3.1.1

Rewrite $\frac{\ln (\frac{1}{2})}{3}$ as $\frac{1}{3} \ln (\frac{1}{2})$ .

$x = \frac{1}{3} \ln (\frac{1}{2}) + \frac{20}{3}$

Step 4.4.3.1.2

Simplify $\frac{1}{3} \ln (\frac{1}{2})$ by moving $\frac{1}{3}$ inside the logarithm.

$x = \ln ({(\frac{1}{2})}^{\frac{1}{3}}) + \frac{20}{3}$

Step 4.4.3.1.3

Apply the product rule to $\frac{1}{2}$ .

$x = \ln (\frac{1^{\frac{1}{3}}}{2^{\frac{1}{3}}}) + \frac{20}{3}$

Step 4.4.3.1.4

One to any power is one.

$x = \ln (\frac{1}{2^{\frac{1}{3}}}) + \frac{20}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \ln (\frac{1}{2^{\frac{1}{3}}}) + \frac{20}{3}$

Decimal Form:

$x = 6.43561760 \dots$