To solve the equation [tex]\(2^{3x} = \frac{\sqrt{2}}{2}\)[/tex], we need to find the value of [tex]\(x\)[/tex]. Here's a detailed, step-by-step solution:
### Step 1: Simplify the Right Side
First, let's simplify the right-hand side of the equation [tex]\(\frac{\sqrt{2}}{2}\)[/tex]. We know that [tex]\(\sqrt{2} = 2^{1/2}\)[/tex], so:
[tex]\[ \frac{\sqrt{2}}{2} = \frac{2^{1/2}}{2} \][/tex]
Since [tex]\(2\)[/tex] can be written as [tex]\(2^1\)[/tex], we have:
[tex]\[ \frac{2^{1/2}}{2^1} = 2^{1/2 - 1} = 2^{-1/2} \][/tex]
Thus, our equation now looks like:
[tex]\[ 2^{3x} = 2^{-1/2} \][/tex]
### Step 2: Equate the Exponents
Now, because the bases are the same, we can equate the exponents:
[tex]\[ 3x = -\frac{1}{2} \][/tex]
### Step 3: Solve the Equation for [tex]\(x\)[/tex]
To find [tex]\(x\)[/tex], we divide both sides of the equation by 3:
[tex]\[ x = -\frac{1}{2} \div 3 = -\frac{1}{6} \][/tex]
So, one of the solutions is:
[tex]\[ x = -\frac{1}{6} \][/tex]
### Step 4: Consider the General Solution Involving Complex Numbers
Next, we should consider the complex solutions. The general solution for an equation of the form [tex]\(2^{3x} = 2^{-1/2}\)[/tex] in the context of complex numbers includes adding multiples of [tex]\(2\pi i\)[/tex] (the period of the exponential function in complex analysis).
Therefore, the general solution can be written as:
[tex]\[ 3x = -\frac{1}{2} + 2k\pi i \log_2(e) \][/tex]
where [tex]\(k\)[/tex] is an integer, and [tex]\( i \)[/tex] is the imaginary unit.
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{-\frac{1}{2} + 2k\pi i \log_2 (e)}{3} \][/tex]
Converting to natural logarithms, we get:
[tex]\[ x = \frac{-\ln(2) + 4k\pi i \ln(e)}{6 \ln(2)} \][/tex]
Since [tex]\(\ln(e)\)[/tex] is 1, it simplifies to:
[tex]\[ x = \frac{-\ln(2) + 4k\pi i}{6 \ln(2)} \][/tex]
For [tex]\( k = 0 \)[/tex], we get the real solution:
[tex]\[ x = \frac{-\frac{1}{2}}{3} = -\frac{1}{6} \][/tex]
For [tex]\( k = 1 \)[/tex], we get:
[tex]\[ x = \frac{-\ln(2) + 4\pi i}{6 \ln(2)} \][/tex]
For [tex]\( k = -1 \)[/tex], we get:
[tex]\[ x = \frac{-\ln(2) - 4\pi i}{6 \ln(2)} \][/tex]
### Step 5: List All Solutions
So, the complete set of solutions is:
[tex]\[ x = -\frac{1}{6} \][/tex]
[tex]\[ x = \frac{-\ln(2) + 4\pi i}{6 \ln(2)} \][/tex]
[tex]\[ x = \frac{-\ln(2) - 4\pi i}{6 \ln(2)} \][/tex]
Thus, the solutions to the equation [tex]\(2^{3x} = \frac{\sqrt{2}}{2}\)[/tex] are:
[tex]\[ x = -\frac{1}{6}, \quad x = \frac{-\ln(2) + 4\pi i}{6 \ln(2)}, \quad x = \frac{-\ln(2) - 4\pi i}{6 \ln(2)} \][/tex]
