Solve for x log base 4 of x-4=- log base 4 of x-7+1

$\log_{4} (x - 4) = - \log_{4} (x - 7) + 1$

Step 1

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

$\log_{4} (x - 4) + \log_{4} (x - 7) = 1$

Step 2

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

$\log_{4} ((x - 4) (x - 7)) = 1$

Step 3

Expand $(x - 4) (x - 7)$ using the FOIL Method.

Apply the distributive property.

$\log_{4} (x (x - 7) - 4 (x - 7)) = 1$

Apply the distributive property.

$\log_{4} (x \cdot x + x \cdot - 7 - 4 (x - 7)) = 1$

Apply the distributive property.

$\log_{4} (x \cdot x + x \cdot - 7 - 4 x - 4 \cdot - 7) = 1$

Step 4

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{4} (x^{2} + x \cdot - 7 - 4 x - 4 \cdot - 7) = 1$

Move $- 7$ to the left of $x$ .

$\log_{4} (x^{2} - 7 \cdot x - 4 x - 4 \cdot - 7) = 1$

Multiply $- 4$ by $- 7$ .

$\log_{4} (x^{2} - 7 x - 4 x + 28) = 1$

Subtract $4 x$ from $- 7 x$ .

$\log_{4} (x^{2} - 11 x + 28) = 1$

Step 5

Rewrite $\log_{4} (x^{2} - 11 x + 28) = 1$ 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 $b^{y} = x$ .

$4^{1} = x^{2} - 11 x + 28$

Step 6

Solve for $x$ .

Rewrite the equation as $x^{2} - 11 x + 28 = 4^{1}$ .

$x^{2} - 11 x + 28 = 4$

Subtract $4$ from both sides of the equation.

$x^{2} - 11 x + 28 - 4 = 0$

Subtract $4$ from $28$ .

$x^{2} - 11 x + 24 = 0$

Factor $x^{2} - 11 x + 24$ 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 $24$ and whose sum is $- 11$ .

$- 8, - 3$

Write the factored form using these integers.

$(x - 8) (x - 3) = 0$

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

$x - 8 = 0$

$x - 3 = 0$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

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.

$x = 3$

The final solution is all the values that make $(x - 8) (x - 3) = 0$ true.

$x = 8, 3$

Step 7

Exclude the solutions that do not make $\log_{4} (x - 4) = - \log_{4} (x - 7) + 1$ true.

$x = 8$

Solve for x log base 4 of x-4=- log base 4 of x-7+1

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