QuestionSolve for x . $\log _{8} x+\log _{8}(x-12)=2$

Studdy Solution

STEP 1

1. The equation involves logarithms with the same base, which allows us to use logarithmic properties to simplify the expression.
2. The domain of the logarithmic function requires that $x > 0$ and $x - 12 > 0$ .

STEP 2

1. Use the properties of logarithms to combine the logarithmic terms.
2. Convert the logarithmic equation to an exponential equation.
3. Solve the resulting quadratic equation for $x$ .
4. Verify the solution(s) to ensure they satisfy the domain restrictions.

STEP 3

Use the product property of logarithms, which states that $\log_b(m) + \log_b(n) = \log_b(mn)$ , to combine the logarithms:
$\log_{8} x + \log_{8} (x - 12) = \log_{8} (x(x - 12))$
Thus, the equation becomes:
$\log_{8} (x^2 - 12x) = 2$

STEP 4

Convert the logarithmic equation to an exponential equation. Recall that if $\log_b(y) = c$ , then $y = b^c$ :
$x^2 - 12x = 8^2$
Calculate $8^2$ :
$x^2 - 12x = 64$

STEP 5

Rearrange the equation to form a standard quadratic equation:
$x^2 - 12x - 64 = 0$
Solve this quadratic equation using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -12$ , and $c = -64$ :
Calculate the discriminant:
$b^2 - 4ac = (-12)^2 - 4(1)(-64) = 144 + 256 = 400$
Calculate the roots:
$x = \frac{-(-12) \pm \sqrt{400}}{2 \times 1} = \frac{12 \pm 20}{2}$
This gives two potential solutions:
$x = \frac{12 + 20}{2} = 16$ $x = \frac{12 - 20}{2} = -4$

STEP 6

Verify the solutions to ensure they satisfy the domain restrictions $x > 0$ and $x - 12 > 0$ :
1. For $x = 16$ : - $x > 0$ is satisfied. - $x - 12 = 4 > 0$ is satisfied.
2. For $x = -4$ : - $x > 0$ is not satisfied.
Thus, the only valid solution is $x = 16$ .
The value of $x$ is:
$\boxed{16}$

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