Solve for x natural log of x+ natural log of x-4 = natural log of 3x-10

$\ln (x) + \ln (x - 4) = \ln (3 x - 10)$

Simplify the left side.

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

$\ln (x (x - 4)) = \ln (3 x - 10)$

Apply the distributive property.

$\ln (x \cdot x + x \cdot - 4) = \ln (3 x - 10)$

Simplify the expression.

Multiply $x$ by $x$ .

$\ln (x^{2} + x \cdot - 4) = \ln (3 x - 10)$

Move $- 4$ to the left of $x$ .

$\ln (x^{2} - 4 x) = \ln (3 x - 10)$

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

$x^{2} - 4 x = 3 x - 10$

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

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

$x^{2} - 4 x - 3 x = - 10$

Subtract $3 x$ from $- 4 x$ .

$x^{2} - 7 x = - 10$

Add $10$ to both sides of the equation.

$x^{2} - 7 x + 10 = 0$

Factor $x^{2} - 7 x + 10$ 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 $- 7$ .

$- 5, - 2$

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 $0$ .

$x - 2 = 0$

Add $2$ to 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 $\ln (x) + \ln (x - 4) = \ln (3 x - 10)$ true.

$x = 5$

Solve for x natural log of x+ natural log of x-4 = natural log of 3x-10

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