To solve the equation [tex]\(0 = -(2)^{x+1} - 2\)[/tex], follow these steps:
1. Isolate the Exponential Term:
Begin by isolating the exponential expression by adding 2 to both sides of the equation:
[tex]\[
0 + 2 = -(2)^{x+1} - 2 + 2
\][/tex]
Simplifying this, we get:
[tex]\[
2 = -(2)^{x+1}
\][/tex]
2. Remove the Negative Sign:
Next, remove the negative sign from the right-hand side by multiplying both sides of the equation by -1:
[tex]\[
-2 = (2)^{x+1}
\][/tex]
3. Express in Terms of Natural Logarithm and Complex Number:
Since the equation involves a negative number on the left-hand side equal to an exponential function with a base greater than 1, it suggests the need to involve complex numbers. Knowing the properties of logarithms and exponentials, we use the base of the logarithm, [tex]\(e\)[/tex], to solve for the exponent [tex]\(x+1\)[/tex].
4. Apply Complex Logarithms:
Recall that [tex]\(e^{i\pi} = -1\)[/tex]. Rewriting, we get:
[tex]\[
-2 = 2 \cdot (-1) = 2 \cdot e^{i\pi}
\][/tex]
Substituting back:
[tex]\[
2^{x+1} = 2 \cdot e^{i\pi}
\][/tex]
5. Take Natural Logarithm:
Take the natural logarithm on both sides of the equation:
[tex]\[
\ln(2^{x+1}) = \ln(2 \cdot e^{i\pi})
\][/tex]
Using properties of logarithms:
[tex]\[
(x+1)\ln(2) = \ln(2) + \ln(e^{i\pi})
\][/tex]
Knowing that [tex]\( \ln(e^{i\pi}) = i\pi \)[/tex]:
[tex]\[
(x+1)\ln(2) = \ln(2) + i\pi
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Isolate [tex]\(x\)[/tex] by subtracting [tex]\(\ln(2)\)[/tex] from both sides:
[tex]\[
x\ln(2) + \ln(2) = \ln(2) + i\pi
\][/tex]
[tex]\[
x\ln(2) = i\pi
\][/tex]
Divide both sides by [tex]\(\ln(2)\)[/tex] to solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{i\pi}{\ln(2)}
\][/tex]
Thus, the solution to the equation [tex]\(0 = -(2)^{x+1} - 2\)[/tex] is:
[tex]\[
x = \frac{i\pi}{\ln(2)}
\][/tex]
