Algebra Examples

$\log_{5} (2 x^{2} - 1) - \log_{5} (3 x + 4) = 0$

Step 1

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

$\log_{5} (\frac{2 x^{2} - 1}{3 x + 4}) = 0$

Step 2

Rewrite $\log_{5} (\frac{2 x^{2} - 1}{3 x + 4}) = 0$ 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 $b^{y} = x$ .

$5^{0} = \frac{2 x^{2} - 1}{3 x + 4}$

Step 3

Cross multiply to remove the fraction.

$2 x^{2} - 1 = 5^{0} (3 x + 4)$

Step 4

Simplify $5^{0} (3 x + 4)$ .

Tap for more steps...

Step 4.1

Anything raised to $0$ is $1$ .

$2 x^{2} - 1 = 1 (3 x + 4)$

Step 4.2

Multiply $3 x + 4$ by $1$ .

$2 x^{2} - 1 = 3 x + 4$

Step 5

Subtract $3 x$ from both sides of the equation.

$2 x^{2} - 1 - 3 x = 4$

Step 6

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

Tap for more steps...

Step 6.1

Add $1$ to both sides of the equation.

$2 x^{2} - 3 x = 4 + 1$

Step 6.2

Add $4$ and $1$ .

$2 x^{2} - 3 x = 5$

Step 7

Subtract $5$ from both sides of the equation.

$2 x^{2} - 3 x - 5 = 0$

Step 8

Factor by grouping.

Tap for more steps...

Step 8.1

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = - 3$ .

Tap for more steps...

Step 8.1.1

Factor $- 3$ out of $- 3 x$ .

$2 x^{2} - 3 x - 5 = 0$

Step 8.1.2

Rewrite $- 3$ as $2$ plus $- 5$

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

Step 8.1.3

Apply the distributive property.

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

Step 8.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 8.2.1

Group the first two terms and the last two terms.

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

Step 8.2.2

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

$2 x (x + 1) - 5 (x + 1) = 0$

Step 8.3

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

$(x + 1) (2 x - 5) = 0$

Step 9

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

$x + 1 = 0$

$2 x - 5 = 0$

Step 10

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

Tap for more steps...

Step 10.1

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

$x + 1 = 0$

Step 10.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 11

Set $2 x - 5$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 11.1

Set $2 x - 5$ equal to $0$ .

$2 x - 5 = 0$

Step 11.2

Solve $2 x - 5 = 0$ for $x$ .

Tap for more steps...

Step 11.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 11.2.2

Divide each term in $2 x = 5$ by $2$ and simplify.

Tap for more steps...

Step 11.2.2.1

Divide each term in $2 x = 5$ by $2$ .

$\frac{2 x}{2} = \frac{5}{2}$

Step 11.2.2.2

Simplify the left side.

Tap for more steps...

Step 11.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 11.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{5}{2}$

Step 11.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{5}{2}$

Step 12

The final solution is all the values that make $(x + 1) (2 x - 5) = 0$ true.

$x = - 1, \frac{5}{2}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = - 1, \frac{5}{2}$

Decimal Form:

$x = - 1, 2.5$

Mixed Number Form:

$x = - 1, 2 \frac{1}{2}$