To solve the equation [tex]\(\frac{1}{2 \times 2^x} = 2^x\)[/tex] step-by-step, follow these detailed steps:
1. Rewrite the equation for clarity:
[tex]\[
\frac{1}{2 \cdot 2^x} = 2^x
\][/tex]
2. Simplify the left-hand side of the equation:
Note that [tex]\(2 \cdot 2^x\)[/tex] can be written as [tex]\(2^{x+1}\)[/tex] because [tex]\(2^x \cdot 2 = 2^{x+1}\)[/tex]. Thus, the equation becomes:
[tex]\[
\frac{1}{2^{x+1}} = 2^x
\][/tex]
3. Rewrite the equation:
We know that [tex]\(\frac{1}{2^{x+1}}\)[/tex] can be expressed as [tex]\((2^{x+1})^{-1}\)[/tex]. Therefore:
[tex]\[
(2^{x+1})^{-1} = 2^x
\][/tex]
This simplifies to:
[tex]\[
2^{-(x+1)} = 2^x
\][/tex]
4. Since the bases are the same (base 2), set the exponents equal to each other:
[tex]\[
-(x+1) = x
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
[tex]\[
-x - 1 = x
\][/tex]
Add [tex]\(x\)[/tex] to both sides:
[tex]\[
-1 = 2x
\][/tex]
Divide both sides by 2:
[tex]\[
x = -\frac{1}{2}
\][/tex]
Thus, we have found one solution: [tex]\( x = -\frac{1}{2} \)[/tex].
6. Consider the comprehensive solution:
Notice that [tex]\(2^x\)[/tex] represents an exponential function which can also involve complex numbers when involving logarithms.
[tex]\[
\frac{1}{2^{x+1}} = 2^x
\][/tex]
If we take the logarithm of both sides, considering natural logarithms:
[tex]\[
\ln\left(\frac{1}{2^{x+1}}\right) = \ln(2^x)
\][/tex]
By the properties of logarithms:
[tex]\[
\ln(1) - \ln(2^{x+1}) = x \ln(2)
\][/tex]
Since [tex]\(\ln(1) = 0\)[/tex], this simplifies to:
[tex]\[
-\ln(2^{x+1}) = x \ln(2)
\][/tex]
Which can be written as:
[tex]\[
-(x+1) \ln(2) = x \ln(2)
\][/tex]
Dividing both sides by [tex]\(\ln(2)\)[/tex]:
[tex]\[
-(x+1) = x
\][/tex]
As previously, solve for [tex]\(x\)[/tex], and we find:
[tex]\[
x = -\frac{1}{2}
\][/tex]
Additionally, there is a complex solution:
[tex]\[
x = -\frac{1}{2} + \frac{i\pi}{\log(2)}
\][/tex]
7. Final solutions:
Therefore, the complete set of solutions to the equation [tex]\(\frac{1}{2 \times 2^x} = 2^x\)[/tex] are:
[tex]\[
x = -\frac{1}{2}, \quad x = -\frac{1}{2} + \frac{i\pi}{\log(2)}
\][/tex]
