Algebra Examples

$\log_{5} (9 x) - \log_{5} (5) = 0$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{5} (\frac{9 x}{5}) = 0$

Step 1.2

Rewrite $\frac{9 x}{5}$ as $9 x \cdot 5^{- 1}$ .

$\log_{5} (9 x \cdot 5^{- 1}) = 0$

Step 1.3

Rewrite $\log_{5} (9 x \cdot 5^{- 1})$ as $\log_{5} (9 x) + \log_{5} (5^{- 1})$ .

$\log_{5} (9 x) + \log_{5} (5^{- 1}) = 0$

Step 1.4

Use logarithm rules to move $- 1$ out of the exponent.

$\log_{5} (9 x) - \log_{5} (5) = 0$

Step 1.5

Logarithm base $5$ of $5$ is $1$ .

$\log_{5} (9 x) - 1 \cdot 1 = 0$

Step 1.6

Multiply $- 1$ by $1$ .

$\log_{5} (9 x) - 1 = 0$

Step 2

Add $1$ to both sides of the equation.

$\log_{5} (9 x) = 1$

Step 3

Rewrite $\log_{5} (9 x) = 1$ 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 $b^{y} = x$ .

$5^{1} = 9 x$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $9 x = 5^{1}$ .

$9 x = 5$

Step 4.2

Divide each term in $9 x = 5^{1}$ by $9$ and simplify.

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Step 4.2.1

Divide each term in $9 x = 5^{1}$ by $9$ .

$\frac{9 x}{9} = \frac{5^{1}}{9}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $9$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{9 x}{9} = \frac{5^{1}}{9}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5^{1}}{9}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Evaluate the exponent.

$x = \frac{5}{9}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{5}{9}$

Decimal Form:

$x = 0.\overline{5}$