Let’s solve the equation [tex]\( \frac{1}{2 \times 2^x} = 2^x \)[/tex].
1. Rewrite the equation:
[tex]\[
\frac{1}{2 \times 2^x} = 2^x
\][/tex]
2. Simplify the left-hand side:
Since [tex]\( 2 \times 2^x \)[/tex] is equal to [tex]\( 2^{x+1} \)[/tex], the equation becomes:
[tex]\[
\frac{1}{2^{x+1}} = 2^x
\][/tex]
3. Express the equation with a common base:
We know that [tex]\( \frac{1}{a} = a^{-1} \)[/tex]. Therefore, the equation [tex]\(\frac{1}{2^{x+1}}\)[/tex] can be rewritten using a negative exponent:
[tex]\[
2^{-(x+1)} = 2^x
\][/tex]
4. Set the exponents equal to each other:
Since the bases (2) are the same, we can set the exponents equal to each other:
[tex]\[
-(x+1) = x
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
-(x+1) = x \implies -x-1 = x \implies -2x = 1 \implies x = -\frac{1}{2}
\][/tex]
6. Check for additional solutions involving complex numbers:
Considering the periodic nature of exponential functions with complex exponents, we should also account for solutions involving complex numbers. When considering the periodic nature of complex exponents, another form of the solution can involve imaginary parts.
This extended solution can be found as:
[tex]\[
x = -\frac{1}{2} + i \cdot \frac{k \pi}{\log 2} \quad \text{for integer values of } k
\][/tex]
For [tex]\( k = 1 \)[/tex]:
[tex]\[
x = -\frac{1}{2} + i \cdot \frac{ \pi}{\log 2}
\][/tex]
Thus, the solutions to the equation [tex]\( \frac{1}{2 \times 2^x} = 2^x \)[/tex] are:
[tex]\[
x = -\frac{1}{2}
\][/tex]
and
[tex]\[
x = -\frac{1}{2} + i \cdot \frac{\pi}{\log 2}
\][/tex]
