Find All Complex Solutions log base 4 of x+ log base 4 of x-3 = log base 4 of -7x+21

$\log_{4} (x) + \log_{4} (x - 3) = \log_{4} (- 7 x + 21)$

Simplify the left side.

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

$\log_{4} (x (x - 3)) = \log_{4} (- 7 x + 21)$

Apply the distributive property.

$\log_{4} (x \cdot x + x \cdot - 3) = \log_{4} (- 7 x + 21)$

Simplify the expression.

Multiply $x$ by $x$ .

$\log_{4} (x^{2} + x \cdot - 3) = \log_{4} (- 7 x + 21)$

Move $- 3$ to the left of $x$ .

$\log_{4} (x^{2} - 3 x) = \log_{4} (- 7 x + 21)$

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

$x^{2} - 3 x = - 7 x + 21$

Solve for $x$ .

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

Add $7 x$ to both sides of the equation.

$x^{2} - 3 x + 7 x = 21$

Add $- 3 x$ and $7 x$ .

$x^{2} + 4 x = 21$

Subtract $21$ from both sides of the equation.

$x^{2} + 4 x - 21 = 0$

Factor $x^{2} + 4 x - 21$ 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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

$(x - 3) (x + 7) = 0$

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

$x - 3 = 0$

$x + 7 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

The final solution is all the values that make $(x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Exclude the solutions that do not make $\log_{4} (x) + \log_{4} (x - 3) = \log_{4} (- 7 x + 21)$ true.

$No solution$

Find All Complex Solutions log base 4 of x+ log base 4 of x-3 = log base 4 of -7x+21

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