Sure, let’s solve the equation [tex]\(2^{x+1} = e^{1-x}\)[/tex] step by step.
1. Rewrite the equation: Start with the original equation.
[tex]\[
2^{x+1} = e^{1-x}
\][/tex]
2. Introduce logarithms: To simplify, we will take the natural logarithm ([tex]\(\ln\)[/tex]) on both sides of the equation. This allows us to use the properties of logarithms to simplify the exponents.
[tex]\[
\ln(2^{x+1}) = \ln(e^{1-x})
\][/tex]
3. Apply the logarithm properties: Use the power rule of logarithms, which states [tex]\(\ln(a^b) = b \ln(a)\)[/tex], on both sides of the equation.
[tex]\[
(x+1)\ln(2) = (1-x)\ln(e)
\][/tex]
4. Simplify the equation: Recall that [tex]\(\ln(e) = 1\)[/tex]. Therefore, the equation simplifies to:
[tex]\[
(x+1)\ln(2) = 1-x
\][/tex]
5. Distribute the logarithms and simplify: Expand and rearrange the terms to isolate [tex]\(x\)[/tex].
[tex]\[
x \ln(2) + \ln(2) = 1 - x
\][/tex]
[tex]\[
x \ln(2) + x = 1 - \ln(2)
\][/tex]
6. Combine like terms: Factor out [tex]\(x\)[/tex] on the left side.
[tex]\[
x(\ln(2) + 1) = 1 - \ln(2)
\][/tex]
7. Solve for [tex]\(x\)[/tex]: Divide both sides by [tex]\((\ln(2) + 1)\)[/tex] to isolate [tex]\(x\)[/tex].
[tex]\[
x = \frac{1 - \ln(2)}{\ln(2) + 1}
\][/tex]
So, the solution to the equation [tex]\(2^{x+1} = e^{1-x}\)[/tex] is:
[tex]\[
x = \frac{1 - \ln(2)}{\ln(2) + 1}
\][/tex]
This is the value of [tex]\(x\)[/tex] that satisfies the given equation.
