Solve for x 2 log of x = log of 2+ log of 3x-4

$2 \log (x) = \log (2) + \log (3 x - 4)$

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

$\log (x^{2}) = \log (2) + \log (3 x - 4)$

Simplify $\log (2) + \log (3 x - 4)$ .

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

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

Apply the distributive property.

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

Multiply.

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Multiply $3$ by $2$ .

$\log (x^{2}) = \log (6 x + 2 \cdot - 4)$

Multiply $2$ by $- 4$ .

$\log (x^{2}) = \log (6 x - 8)$

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

$x^{2} = 6 x - 8$

Solve for $x$ .

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Subtract $6 x$ from both sides of the equation.

$x^{2} - 6 x = - 8$

Move $8$ to the left side of the equation by adding it to both sides.

$x^{2} - 6 x + 8 = 0$

Factor $x^{2} - 6 x + 8$ using the AC method.

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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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

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

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Set the factor equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 2$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The solution is the result of $x - 4 = 0$ and $x - 2 = 0$ .

$x = 4, 2$

Solve for x 2 log of x = log of 2+ log of 3x-4

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