Certainly! Let's solve the equation [tex]\( x^{\ln x - 2} = e^3 \)[/tex] step-by-step.
1. Rewrite the Equation Dfferently:
Our given equation is [tex]\( x^{\ln x - 2} = e^3 \)[/tex].
2. Taking the Natural Logarithm:
To simplify the expression, apply the natural logarithm to both sides of the equation:
[tex]\[ \ln\left(x^{\ln x - 2}\right) = \ln(e^3) \][/tex]
3. Utilize Logarithmic Properties:
Recall that [tex]\( \ln(a^b) = b \cdot \ln(a) \)[/tex] and [tex]\( \ln(e^c) = c \)[/tex]:
[tex]\[ (\ln x - 2) \cdot \ln x = 3 \][/tex]
4. Simplify the Expression:
Distribute [tex]\(\ln x\)[/tex] on the left-hand side:
[tex]\[ (\ln x)^2 - 2 \ln x = 3 \][/tex]
5. Set Up a Quadratic Equation:
Let [tex]\( y = \ln x \)[/tex]. Substituting [tex]\( y \)[/tex] into our expression gives:
[tex]\[ y^2 - 2y = 3 \][/tex]
6. Rewrite in Standard Form:
Rearrange to the standard form of a quadratic equation:
[tex]\[ y^2 - 2y - 3 = 0 \][/tex]
7. Solve the Quadratic Equation:
Factorize or use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ y = \frac{2 \pm \sqrt{4 + 12}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ y = \frac{2 \pm 4}{2} \][/tex]
This gives two solutions:
[tex]\[ y = 3 \quad \text{and} \quad y = -1 \][/tex]
8. Back-Substitute [tex]\( y \)[/tex]:
Recall that [tex]\( y = \ln x \)[/tex], so:
[tex]\[ \ln x = 3 \quad \text{and} \quad \ln x = -1 \][/tex]
9. Solve for [tex]\( x \)[/tex]:
Exponentiate to solve for [tex]\( x \)[/tex]:
[tex]\[ x = e^3 \quad \text{and} \quad x = e^{-1} \][/tex]
10. Final Solutions:
The solutions to the equation [tex]\( x^{\ln x - 2} = e^3 \)[/tex] are:
[tex]\[ x = e^3 \quad \text{and} \quad x = \frac{1}{e} \][/tex]
So, the solutions to the equation [tex]\( x^{\ln x - 2} = e^3 \)[/tex] are [tex]\( x = e^3 \)[/tex] and [tex]\( x = e^{-1} = \frac{1}{e} \)[/tex].
