Precalculus Examples

$\log (4) = 1 + \log (x - 1)$

Step 1

Rewrite the equation as $1 + \log (x - 1) = \log (4)$ .

$1 + \log (x - 1) = \log (4)$

Step 2

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

$\log (x - 1) - \log (4) = - 1$

Step 3

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

$\log (\frac{x - 1}{4}) = - 1$

Step 4

Rewrite $\log (\frac{x - 1}{4}) = - 1$ 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 $b^{y} = x$ .

$10^{- 1} = \frac{x - 1}{4}$

Step 5

Solve for $x$ .

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Step 5.1

Rewrite the equation as $\frac{x - 1}{4} = 10^{- 1}$ .

$\frac{x - 1}{4} = 10^{- 1}$

Step 5.2

Multiply both sides of the equation by $4$ .

$4 \frac{x - 1}{4} = 4 \cdot 10^{- 1}$

Step 5.3

Simplify the left side.

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Step 5.3.1

Cancel the common factor of $4$ .

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Step 5.3.1.1

Cancel the common factor.

$4 \frac{x - 1}{4} = 4 \cdot 10^{- 1}$

Step 5.3.1.2

Rewrite the expression.

$x - 1 = 4 \cdot 10^{- 1}$

Step 5.4

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

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Step 5.4.1

Add $1$ to both sides of the equation.

$x = 4 \cdot 10^{- 1} + 1$

Step 5.4.2

Convert $1$ to scientific notation.

$x = 4 \cdot 10^{- 1} + 1 \cdot 10^{0}$

Step 5.4.3

Move the decimal point in $4$ to the left by $1$ place and increase the power of $10^{- 1}$ by $1$ .

$x = 0.4 \cdot 10^{0} + 1 \cdot 10^{0}$

Step 5.4.4

Factor $10^{0}$ out of $0.4 \cdot 10^{0} + 1 \cdot 10^{0}$ .

$x = (0.4 + 1) \cdot 10^{0}$

Step 5.4.5

Add $0.4$ and $1$ .

$x = 1.4 \cdot 10^{0}$

Step 5.4.6

Convert $1.4 \cdot 10^{0}$ from scientific notation.

$x = 1.4$