Algebra Examples

$\log_{b} (x) + \log_{b} (3) = \log_{b} (27) - 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{b} (x \cdot 3) = \log_{b} (27) - 1$

Step 1.2

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

$\log_{b} (3 x) = \log_{b} (27) - 1$

Step 2

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

$\log_{b} (3 x) - \log_{b} (27) = - 1$

Step 3

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

$\log_{b} (\frac{3 x}{27}) = - 1$

Step 4

Cancel the common factor of $3$ and $27$ .

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Step 4.1

Factor $3$ out of $3 x$ .

$\log_{b} (\frac{3 (x)}{27}) = - 1$

Step 4.2

Cancel the common factors.

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Step 4.2.1

Factor $3$ out of $27$ .

$\log_{b} (\frac{3 x}{3 \cdot 9}) = - 1$

Step 4.2.2

Cancel the common factor.

$\log_{b} (\frac{3 x}{3 \cdot 9}) = - 1$

Step 4.2.3

Rewrite the expression.

$\log_{b} (\frac{x}{9}) = - 1$

Step 5

Rewrite $\log_{b} (\frac{x}{9}) = - 1$ 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 $b^{y} = x$ .

$b^{- 1} = \frac{x}{9}$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $\frac{x}{9} = b^{- 1}$ .

$\frac{x}{9} = b^{- 1}$

Step 6.2

Multiply both sides of the equation by $9$ .

$9 \frac{x}{9} = 9 b^{- 1}$

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Cancel the common factor of $9$ .

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Step 6.3.1.1.1

Cancel the common factor.

$9 \frac{x}{9} = 9 b^{- 1}$

Step 6.3.1.1.2

Rewrite the expression.

$x = 9 b^{- 1}$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Simplify $9 b^{- 1}$ .

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Step 6.3.2.1.1

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

$x = 9 \frac{1}{b}$

Step 6.3.2.1.2

Combine $9$ and $\frac{1}{b}$ .

$x = \frac{9}{b}$