To solve the logarithmic equation
[tex]\[
\log_8 x + \log_8 (x+2) = \log_8 (x+56)
\][/tex]
we can make use of the properties of logarithms. Here, we will work through combining logs and solving the resulting equation step-by-step.
### Step 1: Combine the logarithms on the left-hand side
Using the property of logarithms that states [tex]\(\log_b a + \log_b c = \log_b(ac)\)[/tex], we can combine the logs on the left-hand side:
[tex]\[
\log_8 (x) + \log_8 (x+2) = \log_8 (x(x+2))
\][/tex]
So, the equation becomes:
[tex]\[
\log_8 (x(x+2)) = \log_8 (x+56)
\][/tex]
### Step 2: Eliminate the logarithms
Since the bases of the logarithms on both sides are the same (base 8), we can equate the arguments of the logarithms:
[tex]\[
x(x+2) = x + 56
\][/tex]
### Step 3: Simplify the equation
First, expand and simplify the left-hand side:
[tex]\[
x^2 + 2x = x + 56
\][/tex]
Next, move all terms to one side of the equation to form a quadratic equation:
[tex]\[
x^2 + 2x - x - 56 = 0
\][/tex]
[tex]\[
x^2 + x - 56 = 0
\][/tex]
### Step 4: Solve the quadratic equation
We solve the quadratic equation [tex]\(x^2 + x - 56 = 0\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -56\)[/tex].
Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 1^2 - 4(1)(-56) = 1 + 224 = 225
\][/tex]
Calculate the roots:
[tex]\[
x = \frac{-1 \pm \sqrt{225}}{2} = \frac{-1 \pm 15}{2}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{-1 + 15}{2} = \frac{14}{2} = 7
\][/tex]
[tex]\[
x = \frac{-1 - 15}{2} = \frac{-16}{2} = -8
\][/tex]
### Step 5: Check the validity of the solutions
Since the arguments of logarithms must be positive, we need to discard any solutions that make the argument of any logarithm equal to zero or negative.
Checking [tex]\(x = 7\)[/tex]:
[tex]\[
x > 0
\][/tex]
[tex]\[
x + 2 = 7 + 2 = 9 > 0
\][/tex]
[tex]\[
x + 56 = 7 + 56 = 63 > 0
\][/tex]
Hence, [tex]\(x = 7\)[/tex] is valid.
Checking [tex]\(x = -8\)[/tex]:
[tex]\[
x < 0
\][/tex]
Since [tex]\(x < 0\)[/tex] is invalid for logarithmic arguments, we discard [tex]\(x = -8\)[/tex].
### Conclusion
The valid solution, approximated to three decimal places, is:
[tex]\[
x = 7
\][/tex]
