Solve for x log base 4 of x-5+ log base 4 of x-5=1

$\log_{4} (x - 5) + \log_{4} (x - 5) = 1$

Step 1

Simplify the left side.

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

$\log_{4} ((x - 5) (x - 5)) = 1$

Expand $(x - 5) (x - 5)$ using the FOIL Method.

Apply the distributive property.

$\log_{4} (x (x - 5) - 5 (x - 5)) = 1$

Apply the distributive property.

$\log_{4} (x \cdot x + x \cdot - 5 - 5 (x - 5)) = 1$

Apply the distributive property.

$\log_{4} (x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5) = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\log_{4} (x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5) = 1$

Move $- 5$ to the left of $x$ .

$\log_{4} (x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5) = 1$

Multiply $- 5$ by $- 5$ .

$\log_{4} (x^{2} - 5 x - 5 x + 25) = 1$

Subtract $5 x$ from $- 5 x$ .

$\log_{4} (x^{2} - 10 x + 25) = 1$

Step 2

Rewrite $\log_{4} (x^{2} - 10 x + 25) = 1$ 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 $b^{y} = x$ .

$4^{1} = x^{2} - 10 x + 25$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - 10 x + 25 = 4^{1}$ .

$x^{2} - 10 x + 25 = 4$

Subtract $4$ from both sides of the equation.

$x^{2} - 10 x + 25 - 4 = 0$

Subtract $4$ from $25$ .

$x^{2} - 10 x + 21 = 0$

Factor $x^{2} - 10 x + 21$ 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 $- 10$ .

$- 7, - 3$

Write the factored form using these integers.

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

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

$x - 7 = 0$

$x - 3 = 0$

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

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

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

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 $(x - 7) (x - 3) = 0$ true.

$x = 7, 3$

Step 4

Exclude the solutions that do not make $\log_{4} (x - 5) + \log_{4} (x - 5) = 1$ true.

$x = 7$

Solve for x log base 4 of x-5+ log base 4 of x-5=1

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