Solve for x log of x+ log of x-99=2

$\log (x) + \log (x - 99) = 2$

Step 1

Simplify the left side.

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

$\log (x (x - 99)) = 2$

Apply the distributive property.

$\log (x \cdot x + x \cdot - 99) = 2$

Simplify the expression.

Multiply $x$ by $x$ .

$\log (x^{2} + x \cdot - 99) = 2$

Move $- 99$ to the left of $x$ .

$\log (x^{2} - 99 x) = 2$

Step 2

Rewrite $\log (x^{2} - 99 x) = 2$ 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 $b^{y} = x$ .

$10^{2} = x^{2} - 99 x$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 99 x = 10^{2}$ .

$x^{2} - 99 x = 10^{2}$

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

$x^{2} - 99 x = 100$

Subtract $100$ from both sides of the equation.

$x^{2} - 99 x - 100 = 0$

Factor $x^{2} - 99 x - 100$ 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 $- 100$ and whose sum is $- 99$ .

$- 100, 1$

Write the factored form using these integers.

$(x - 100) (x + 1) = 0$

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

$x - 100 = 0$

$x + 1 = 0$

Set $x - 100$ equal to $0$ and solve for $x$ .

Set $x - 100$ equal to $0$ .

$x - 100 = 0$

Add $100$ to both sides of the equation.

$x = 100$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 100) (x + 1) = 0$ true.

$x = 100, - 1$

Step 4

Exclude the solutions that do not make $\log (x) + \log (x - 99) = 2$ true.

$x = 100$

Solve for x log of x+ log of x-99=2

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