Sure! Let's solve the equation step-by-step:
Given equation:
[tex]\[
\frac{1}{2 \times 2^x} = 2^x
\][/tex]
1. Rewrite the given equation for clarity:
The equation is:
[tex]\[
\frac{1}{2 \times 2^x} = 2^x
\][/tex]
2. Multiply both sides by [tex]\(2 \times 2^x\)[/tex] to eliminate the fraction:
[tex]\[
1 = 2 \times 2^x \times 2^x
\][/tex]
3. Simplify the right-hand side using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex] of exponents:
[tex]\[
1 = 2 \times 2^{2x}
\][/tex]
4. Divide both sides of the equation by 2:
[tex]\[
\frac{1}{2} = 2^{2x}
\][/tex]
5. To solve for [tex]\(2x\)[/tex], we'll take the natural logarithm (ln) of both sides, since logarithms allow us to bring exponents down:
[tex]\[
\ln\left(\frac{1}{2}\right) = \ln(2^{2x})
\][/tex]
6. Utilize the property [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[
\ln\left(\frac{1}{2}\right) = 2x \ln(2)
\][/tex]
7. Simplify [tex]\(\ln\left(\frac{1}{2}\right)\)[/tex]:
[tex]\[
\ln\left(\frac{1}{2}\right) = \ln(1) - \ln(2)
\][/tex]
[tex]\[
\ln\left(\frac{1}{2}\right) = 0 - \ln(2)
\][/tex]
[tex]\[
\ln\left(\frac{1}{2}\right) = -\ln(2)
\][/tex]
8. So, we have:
[tex]\[
-\ln(2) = 2x \ln(2)
\][/tex]
9. Solve for [tex]\(x\)[/tex]:
[tex]\[
-1 = 2x
\][/tex]
[tex]\[
x = -\frac{1}{2}
\][/tex]
10. Consider complex solutions:
The logarithm has multiple values due to periodicity in the complex plane. The general solution involves adding multiples of [tex]\(2\pi i\)[/tex]. Hence, the general solution can be expressed as:
[tex]\[
2x \ln(2) = -\ln(2) + 2\pi in
\][/tex]
where [tex]\(n\)[/tex] is any integer.
11. Divide by [tex]\(2 \ln(2)\)[/tex]:
[tex]\[
x = -\frac{1}{2} + \frac{\pi i n}{2 \ln(2)}
\][/tex]
For [tex]\(n=0\)[/tex], we get:
[tex]\[
x = -\frac{1}{2}
\][/tex]
For [tex]\(n=1\)[/tex], we get:
[tex]\[
x = -\frac{1}{2} + \frac{\pi i}{\ln(2)}
\][/tex]
Thus, the solutions to the equation [tex]\(\frac{1}{2 \times 2^x} = 2^x\)[/tex] are:
[tex]\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{1}{2} + \frac{\pi i}{\ln(2)}
\][/tex]
