QuestionSolve $\log _{2} x + \log _{2}(x-2) = 3$ for $x$ . Options: a) $4,-2$ b) 4 c) -2 d) $2,-4$ e) None.

Studdy Solution

STEP 1

Assumptions1. We are given the equation $\log{} x+\log{}(x-)=3$ . We are looking for the value of $x$
3. We assume that the logarithm base is4. We assume that the logarithm properties hold true

STEP 2

We can use the property of logarithms that states that the sum of two logarithms with the same base is equal to the logarithm of the product of the numbers. So we can rewrite the equation as $\log{2} (x \cdot (x-2)) =$

STEP 3

Now, we can use the property of logarithms that states that $\log{b} a = c$ is equivalent to $b^c = a$ . So we can rewrite the equation as $2^3 = x \cdot (x-2)$

STEP 4

Calculate the value of $2^3$ .
$2^3 =8$ So the equation becomes $8 = x \cdot (x-2)$

STEP 5

Expand the right side of the equation.
$8 = x^2 -2x$

STEP 6

Rearrange the equation to form a quadratic equation.
$x^2 -2x -8 =0$

STEP 7

Now, we can solve the quadratic equation. The quadratic equation is of the form $ax^2 + bx + c =0$ and the solution for $x$ is given by the formula $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

STEP 8

Plug in the values for $a$ , $b$ and $c$ into the formula.
$x = \frac{-(-2) \pm \sqrt{(-2)^2 -4 \cdot1 \cdot -8}}{2 \cdot1}$

STEP 9

implify the equation to find the values of $x$ .
$x = \frac{2 \pm \sqrt{4 +32}}{2}$

STEP 10

Calculate the values under the square root.
$x = \frac{2 \pm \sqrt{36}}{2}$

STEP 11

Calculate the square root.
$x = \frac{ \pm6}{}$

STEP 12

Calculate the two possible values for $x$ .
$x =4, -2$ So, the solution is $x =4, -2$ .

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