Algebra Examples

$\ln (x - \frac{7}{2}) + \ln (14) = 2 \ln (x)$

Step 1

Simplify the left side.

Tap for more steps...

Step 1.1

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

$\ln ((x - \frac{7}{2}) \cdot 14) = 2 \ln (x)$

Step 1.2

Apply the distributive property.

$\ln (x \cdot 14 - \frac{7}{2} \cdot 14) = 2 \ln (x)$

Step 1.3

Move $14$ to the left of $x$ .

$\ln (14 \cdot x - \frac{7}{2} \cdot 14) = 2 \ln (x)$

Step 1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 1.4.1

Move the leading negative in $- \frac{7}{2}$ into the numerator.

$\ln (14 \cdot x + \frac{- 7}{2} \cdot 14) = 2 \ln (x)$

Step 1.4.2

Factor $2$ out of $14$ .

$\ln (14 \cdot x + \frac{- 7}{2} \cdot (2 (7))) = 2 \ln (x)$

Step 1.4.3

Cancel the common factor.

$\ln (14 \cdot x + \frac{- 7}{2} \cdot (2 \cdot 7)) = 2 \ln (x)$

Step 1.4.4

Rewrite the expression.

$\ln (14 \cdot x - 7 \cdot 7) = 2 \ln (x)$

Step 1.5

Multiply $- 7$ by $7$ .

$\ln (14 x - 49) = 2 \ln (x)$

Step 2

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$\ln (14 x - 49) = \ln (x^{2})$

Step 3

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$14 x - 49 = x^{2}$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

Subtract $x^{2}$ from both sides of the equation.

$14 x - 49 - x^{2} = 0$

Step 4.2

Factor the left side of the equation.

Tap for more steps...

Step 4.2.1

Factor $- 1$ out of $14 x - 49 - x^{2}$ .

Tap for more steps...

Step 4.2.1.1

Reorder the expression.

Tap for more steps...

Step 4.2.1.1.1

Move $- 49$ .

$14 x - x^{2} - 49 = 0$

Step 4.2.1.1.2

Reorder $14 x$ and $- x^{2}$ .

$- x^{2} + 14 x - 49 = 0$

Step 4.2.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 14 x - 49 = 0$

Step 4.2.1.3

Factor $- 1$ out of $14 x$ .

$- (x^{2}) - (- 14 x) - 49 = 0$

Step 4.2.1.4

Rewrite $- 49$ as $- 1 (49)$ .

$- (x^{2}) - (- 14 x) - 1 \cdot 49 = 0$

Step 4.2.1.5

Factor $- 1$ out of $- (x^{2}) - (- 14 x)$ .

$- (x^{2} - 14 x) - 1 \cdot 49 = 0$

Step 4.2.1.6

Factor $- 1$ out of $- (x^{2} - 14 x) - 1 (49)$ .

$- (x^{2} - 14 x + 49) = 0$

Step 4.2.2

Factor using the perfect square rule.

Tap for more steps...

Step 4.2.2.1

Rewrite $49$ as $7^{2}$ .

$- (x^{2} - 14 x + 7^{2}) = 0$

Step 4.2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$14 x = 2 \cdot x \cdot 7$

Step 4.2.2.3

Rewrite the polynomial.

$- (x^{2} - 2 \cdot x \cdot 7 + 7^{2}) = 0$

Step 4.2.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 7$ .

$- {(x - 7)}^{2} = 0$

Step 4.3

Divide each term in $- {(x - 7)}^{2} = 0$ by $- 1$ and simplify.

Tap for more steps...

Step 4.3.1

Divide each term in $- {(x - 7)}^{2} = 0$ by $- 1$ .

$\frac{- {(x - 7)}^{2}}{- 1} = \frac{0}{- 1}$

Step 4.3.2

Simplify the left side.

Tap for more steps...

Step 4.3.2.1

Dividing two negative values results in a positive value.

$\frac{{(x - 7)}^{2}}{1} = \frac{0}{- 1}$

Step 4.3.2.2

Divide ${(x - 7)}^{2}$ by $1$ .

${(x - 7)}^{2} = \frac{0}{- 1}$

Step 4.3.3

Simplify the right side.

Tap for more steps...

Step 4.3.3.1

Divide $0$ by $- 1$ .

${(x - 7)}^{2} = 0$

Step 4.4

Set the $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 4.5

Add $7$ to both sides of the equation.

$x = 7$