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To solve the equation [tex]\( e^x - 12 e^{-x} - 1 = 0 \)[/tex], we can follow these steps:
### Step 1: Introduce a substitution
Let's make a substitution to transform this equation into a simpler form. Set [tex]\( y = e^x \)[/tex]. Since [tex]\( e^{-x} \)[/tex] is the reciprocal of [tex]\( e^x \)[/tex], we have:
[tex]\[ e^{-x} = \frac{1}{y} \][/tex]
### Step 2: Substitute and simplify
Substitute [tex]\( y \)[/tex] and [tex]\( e^{-x} \)[/tex] into the original equation:
[tex]\[ y - 12 \left(\frac{1}{y}\right) - 1 = 0 \][/tex]
Multiply every term by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ y^2 - 12 - y = 0 \][/tex]
### Step 3: Rearrange into a standard quadratic form
Rearrange the equation to form a quadratic equation:
[tex]\[ y^2 - y - 12 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve the quadratic equation [tex]\( y^2 - y - 12 = 0 \)[/tex], we can factorize it. We look for two numbers that multiply to [tex]\(-12\)[/tex] and add up to [tex]\(-1\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(3\)[/tex]. Therefore, we can write:
[tex]\[ (y - 4)(y + 3) = 0 \][/tex]
Set each factor equal to zero and solve for [tex]\( y \)[/tex]:
[tex]\[ y - 4 = 0 \quad \text{or} \quad y + 3 = 0 \][/tex]
[tex]\[ y = 4 \quad \text{or} \quad y = -3 \][/tex]
### Step 5: Back-substitute and solve for [tex]\( x \)[/tex]
Recall that [tex]\( y = e^x \)[/tex]. We now substitute back to find [tex]\( x \)[/tex]:
1. For [tex]\( y = 4 \)[/tex]:
[tex]\[ e^x = 4 \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ x = \ln 4 \][/tex]
2. For [tex]\( y = -3 \)[/tex]:
[tex]\[ e^x = -3 \][/tex]
However, the exponential function [tex]\( e^x \)[/tex] is always positive, so [tex]\( e^x = -3 \)[/tex] has no real solution. But there could be complex solutions:
For complex solutions, we rewrite [tex]\(-3\)[/tex] as [tex]\(3 e^{i (\pi + 2k\pi)}\)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex]:
[tex]\[ e^x = 3 e^{i (\pi + 2k\pi)} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ x = \ln 3 + i(\pi + 2k\pi) \][/tex]
When [tex]\( k = 0 \)[/tex], this simplifies to:
[tex]\[ x = \ln 3 + i\pi \][/tex]
Thus, the two solutions for the initial equation are:
[tex]\[ x = \ln 4 \][/tex]
and
[tex]\[ x = \ln 3 + i\pi \][/tex]
### Step 1: Introduce a substitution
Let's make a substitution to transform this equation into a simpler form. Set [tex]\( y = e^x \)[/tex]. Since [tex]\( e^{-x} \)[/tex] is the reciprocal of [tex]\( e^x \)[/tex], we have:
[tex]\[ e^{-x} = \frac{1}{y} \][/tex]
### Step 2: Substitute and simplify
Substitute [tex]\( y \)[/tex] and [tex]\( e^{-x} \)[/tex] into the original equation:
[tex]\[ y - 12 \left(\frac{1}{y}\right) - 1 = 0 \][/tex]
Multiply every term by [tex]\( y \)[/tex] to clear the fraction:
[tex]\[ y^2 - 12 - y = 0 \][/tex]
### Step 3: Rearrange into a standard quadratic form
Rearrange the equation to form a quadratic equation:
[tex]\[ y^2 - y - 12 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve the quadratic equation [tex]\( y^2 - y - 12 = 0 \)[/tex], we can factorize it. We look for two numbers that multiply to [tex]\(-12\)[/tex] and add up to [tex]\(-1\)[/tex]. These numbers are [tex]\(-4\)[/tex] and [tex]\(3\)[/tex]. Therefore, we can write:
[tex]\[ (y - 4)(y + 3) = 0 \][/tex]
Set each factor equal to zero and solve for [tex]\( y \)[/tex]:
[tex]\[ y - 4 = 0 \quad \text{or} \quad y + 3 = 0 \][/tex]
[tex]\[ y = 4 \quad \text{or} \quad y = -3 \][/tex]
### Step 5: Back-substitute and solve for [tex]\( x \)[/tex]
Recall that [tex]\( y = e^x \)[/tex]. We now substitute back to find [tex]\( x \)[/tex]:
1. For [tex]\( y = 4 \)[/tex]:
[tex]\[ e^x = 4 \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ x = \ln 4 \][/tex]
2. For [tex]\( y = -3 \)[/tex]:
[tex]\[ e^x = -3 \][/tex]
However, the exponential function [tex]\( e^x \)[/tex] is always positive, so [tex]\( e^x = -3 \)[/tex] has no real solution. But there could be complex solutions:
For complex solutions, we rewrite [tex]\(-3\)[/tex] as [tex]\(3 e^{i (\pi + 2k\pi)}\)[/tex] for [tex]\( k \in \mathbb{Z} \)[/tex]:
[tex]\[ e^x = 3 e^{i (\pi + 2k\pi)} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ x = \ln 3 + i(\pi + 2k\pi) \][/tex]
When [tex]\( k = 0 \)[/tex], this simplifies to:
[tex]\[ x = \ln 3 + i\pi \][/tex]
Thus, the two solutions for the initial equation are:
[tex]\[ x = \ln 4 \][/tex]
and
[tex]\[ x = \ln 3 + i\pi \][/tex]
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