To solve the equation [tex]\( 2^{-3x} = 2^{x-1} \)[/tex], we will utilize properties of exponents and logarithms. Here is the step-by-step process:
1. Equate the exponents:
Since the bases on both sides of the equation are the same (both are [tex]\(2\)[/tex]), we can set the exponents equal to each other. Therefore, we have:
[tex]\[
-3x = x - 1
\][/tex]
2. Solve for [tex]\( x \)[/tex]:
To isolate [tex]\( x \)[/tex], we move all the [tex]\( x \)[/tex]-terms to one side of the equation:
[tex]\[
-3x - x = -1
\][/tex]
Simplify the left side:
[tex]\[
-4x = -1
\][/tex]
Now, divide both sides by [tex]\(-4\)[/tex]:
[tex]\[
x = \frac{1}{4}
\][/tex]
3. Check for additional complex solutions:
In more general terms, this equation may have complex solutions due to the properties of logarithms and complex numbers. To find all possible solutions, we recognize that the exponentiation of complex logarithms can lead to additional solutions based on the periodicity of the complex exponential function.
4. Solutions involving complex numbers:
The solutions to the equation [tex]\( 2^{-3x} = 2^{x-1} \)[/tex] can be more formally written, considering periodicity influenced by the factor [tex]\(2\pi i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit and [tex]\( \log(2) \)[/tex] is the natural logarithm of 2. Given these considerations, the complete set of solutions is:
[tex]\[
x = \frac{1}{4}, \quad x = \frac{1}{4} - \frac{i\pi}{2\log(2)}, \quad x = \frac{1}{4} + \frac{i\pi}{2\log(2)}, \quad x = \frac{1}{4} + \frac{i\pi}{\log(2)}
\][/tex]
In summary, the solutions to the equation [tex]\( 2^{-3x} = 2^{x-1} \)[/tex] are:
[tex]\[
x = \frac{1}{4}, \quad x = \frac{1}{4} - \frac{i\pi}{2\log(2)}, \quad x = \frac{1}{4} + \frac{i\pi}{2\log(2)}, \quad x = \frac{1}{4} + \frac{i\pi}{\log(2)}
\][/tex]
