Solve for x log of 5x+ log of x-1=2

$\log (5 x) + \log (x - 1) = 2$

Step 1

Simplify the left side.

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

$\log (5 x (x - 1)) = 2$

Apply the distributive property.

$\log (5 x \cdot x + 5 x \cdot - 1) = 2$

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

Move $x$ .

$\log (5 (x \cdot x) + 5 x \cdot - 1) = 2$

Multiply $x$ by $x$ .

$\log (5 x^{2} + 5 x \cdot - 1) = 2$

Multiply $- 1$ by $5$ .

$\log (5 x^{2} - 5 x) = 2$

Step 2

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

$10^{2} = 5 x^{2} - 5 x$

Step 3

Solve for $x$ .

Rewrite the equation as $5 x^{2} - 5 x = 10^{2}$ .

$5 x^{2} - 5 x = 10^{2}$

Raise $10$ to the power of $2$ .

$5 x^{2} - 5 x = 100$

Subtract $100$ from both sides of the equation.

$5 x^{2} - 5 x - 100 = 0$

Factor the left side of the equation.

Factor $5$ out of $5 x^{2} - 5 x - 100$ .

Factor $5$ out of $5 x^{2}$ .

$5 (x^{2}) - 5 x - 100 = 0$

Factor $5$ out of $- 5 x$ .

$5 (x^{2}) + 5 (- x) - 100 = 0$

Factor $5$ out of $- 100$ .

$5 (x^{2}) + 5 (- x) + 5 (- 20) = 0$

Factor $5$ out of $5 (x^{2}) + 5 (- x)$ .

$5 (x^{2} - x) + 5 (- 20) = 0$

Factor $5$ out of $5 (x^{2} - x) + 5 (- 20)$ .

$5 (x^{2} - x - 20) = 0$

Factor.

Factor $x^{2} - x - 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $- 1$ .

$- 5, 4$

Write the factored form using these integers.

$5 ((x - 5) (x + 4)) = 0$

Remove unnecessary parentheses.

$5 (x - 5) (x + 4) = 0$

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

$x - 5 = 0$

$x + 4 = 0$

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

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

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $5 (x - 5) (x + 4) = 0$ true.

$x = 5, - 4$

Step 4

Exclude the solutions that do not make $\log (5 x) + \log (x - 1) = 2$ true.

$x = 5$

Solve for x log of 5x+ log of x-1=2

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