Let's solve the given exponential equation step by step:
The given equation is:
[tex]\[ e^{3x - 4} \cdot e^{-x} = 2e \][/tex]
First, we simplify the left-hand side. We know that when multiplying exponential expressions with the same base, we add the exponents:
[tex]\[ e^{3x - 4} \cdot e^{-x} = e^{(3x - 4) + (-x)} = e^{3x - 4 - x} = e^{2x - 4} \][/tex]
Now, our equation simplifies to:
[tex]\[ e^{2x - 4} = 2e \][/tex]
Next, we want to isolate the exponential expression. Since the right-hand side has an [tex]\(e\)[/tex], we can divide both sides by [tex]\(e\)[/tex]:
[tex]\[ e^{2x - 4} = 2e \][/tex]
[tex]\[ e^{2x - 4} / e = 2e / e \][/tex]
[tex]\[ e^{2x - 4 - 1} = 2 \][/tex]
[tex]\[ e^{2x - 5} = 2 \][/tex]
To solve for [tex]\(x\)[/tex], we take the natural logarithm (ln) of both sides of the equation. Applying the natural logarithm to both sides, we get:
[tex]\[ \ln(e^{2x - 5}) = \ln(2) \][/tex]
Using the property of logarithms that [tex]\(\ln(e^y) = y\)[/tex], we simplify the left-hand side:
[tex]\[ 2x - 5 = \ln(2) \][/tex]
Now, we solve for [tex]\(x\)[/tex] by isolating it on one side of the equation. First, add 5 to both sides:
[tex]\[ 2x = \ln(2) + 5 \][/tex]
Then, divide both sides by 2:
[tex]\[ x = \frac{\ln(2) + 5}{2} \][/tex]
We know that [tex]\(\ln(2) \approx 0.693\)[/tex]:
[tex]\[ x = \frac{0.693 + 5}{2} = \frac{5.693}{2} \approx 2.847 \][/tex]
Therefore, the solution to the equation, correct to the nearest thousandth, is:
[tex]\[ x \approx 2.847 \][/tex]
