Solve for x log of x-3=1- log of x

$\log (x - 3) = 1 - \log (x)$

Move all the terms containing a logarithm to the left side of the equation.

$\log (x - 3) + \log (x) = 1$

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

$\log ((x - 3) x) = 1$

Apply the distributive property.

$\log (x \cdot x - 3 x) = 1$

Multiply $x$ by $x$ .

$\log (x^{2} - 3 x) = 1$

Rewrite $\log (x^{2} - 3 x) = 1$ 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 $b^{y} = x$ .

$10^{1} = x^{2} - 3 x$

Solve for $x$ .

Rewrite the equation as $x^{2} - 3 x = 10^{1}$ .

$x^{2} - 3 x = 10$

Subtract $10$ from both sides of the equation.

$x^{2} - 3 x - 10 = 0$

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

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

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

$x - 5 = 0$

$x + 2 = 0$

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 $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = 5, - 2$

Exclude the solutions that do not make $\log (x - 3) = 1 - \log (x)$ true.

$x = 5$

Solve for x log of x-3=1- log of x

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