Algebra Examples

$\log_{6} (x) + \log_{6} (5 x - 1) = 1$

Step 1

Simplify the left side.

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Step 1.1

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

$\log_{6} (x (5 x - 1)) = 1$

Step 1.2

Simplify by multiplying through.

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Step 1.2.1

Apply the distributive property.

$\log_{6} (x (5 x) + x \cdot - 1) = 1$

Step 1.2.2

Reorder.

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Step 1.2.2.1

Rewrite using the commutative property of multiplication.

$\log_{6} (5 x \cdot x + x \cdot - 1) = 1$

Step 1.2.2.2

Move $- 1$ to the left of $x$ .

$\log_{6} (5 x \cdot x - 1 \cdot x) = 1$

Step 1.3

Simplify each term.

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Step 1.3.1

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

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Step 1.3.1.1

Move $x$ .

$\log_{6} (5 (x \cdot x) - 1 \cdot x) = 1$

Step 1.3.1.2

Multiply $x$ by $x$ .

$\log_{6} (5 x^{2} - 1 \cdot x) = 1$

Step 1.3.2

Rewrite $- 1 x$ as $- x$ .

$\log_{6} (5 x^{2} - x) = 1$

Step 2

Rewrite $\log_{6} (5 x^{2} - x) = 1$ 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 $b^{y} = x$ .

$6^{1} = 5 x^{2} - x$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $5 x^{2} - x = 6^{1}$ .

$5 x^{2} - x = 6$

Step 3.2

Subtract $6$ from both sides of the equation.

$5 x^{2} - x - 6 = 0$

Step 3.3

Factor by grouping.

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Step 3.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 5 \cdot - 6 = - 30$ and whose sum is $b = - 1$ .

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Step 3.3.1.1

Factor $- 1$ out of $- x$ .

$5 x^{2} - (x) - 6 = 0$

Step 3.3.1.2

Rewrite $- 1$ as $5$ plus $- 6$

$5 x^{2} + (5 - 6) x - 6 = 0$

Step 3.3.1.3

Apply the distributive property.

$5 x^{2} + 5 x - 6 x - 6 = 0$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(5 x^{2} + 5 x) - 6 x - 6 = 0$

Step 3.3.2.2

Factor out the greatest common factor (GCF) from each group.

$5 x (x + 1) - 6 (x + 1) = 0$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (5 x - 6) = 0$

Step 3.4

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

$x + 1 = 0$

$5 x - 6 = 0$

Step 3.5

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

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Step 3.5.1

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

$x + 1 = 0$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

Set $5 x - 6$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $5 x - 6$ equal to $0$ .

$5 x - 6 = 0$

Step 3.6.2

Solve $5 x - 6 = 0$ for $x$ .

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Step 3.6.2.1

Add $6$ to both sides of the equation.

$5 x = 6$

Step 3.6.2.2

Divide each term in $5 x = 6$ by $5$ and simplify.

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Step 3.6.2.2.1

Divide each term in $5 x = 6$ by $5$ .

$\frac{5 x}{5} = \frac{6}{5}$

Step 3.6.2.2.2

Simplify the left side.

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Step 3.6.2.2.2.1

Cancel the common factor of $5$ .

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Step 3.6.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{6}{5}$

Step 3.6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{6}{5}$

Step 3.7

The final solution is all the values that make $(x + 1) (5 x - 6) = 0$ true.

$x = - 1, \frac{6}{5}$

Step 4

Exclude the solutions that do not make $\log_{6} (x) + \log_{6} (5 x - 1) = 1$ true.

$x = \frac{6}{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{6}{5}$

Decimal Form:

$x = 1.2$

Mixed Number Form:

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