Using the logarithmic identity [tex]\ln(a^b) = b\ln(a)[/tex], we can simplify this to:
[tex]-x^x \ln(x) = \sqrt{2} \ln(2)[/tex]
Dividing both sides by [tex]-\ln(x)[/tex], we get:
[tex]x^x = -\frac{\sqrt{2}\ln(2)}{\ln(x)}[/tex]
Now, we can use the Lambert W function, which is defined as the inverse function of [tex]f(z) = ze^z[/tex], to solve for x. Applying the function to both sides, we get:
Note that the argument of the Lambert W function is [tex]-\frac{\sqrt{2}\ln(2)}{x}[/tex], which is a constant. Therefore, we can simplify the equation further by substituting [tex]t = -\frac{\sqrt{2}\ln(2)}{x}[/tex], so that:
[tex]x = -\frac{\ln(2)}{W(t)}[/tex]
[tex]t = -\frac{\sqrt{2}\ln(2)}{x}[/tex]
Solving for t, we get:
[tex]t = -\frac{\sqrt{2}\ln(2)}{x} \implies x = -\frac{\sqrt{2}\ln(2)}{t}[/tex]
Substituting this expression for x into the original equation [tex]x^{-x^x} = 2^{\sqrt{2}}[/tex], we get:
This equation cannot be solved algebraically, but it can be approximated numerically using a computer or calculator. Using a numerical solver, we can find that:
[tex]t \approx -1.02938[/tex]
Substituting this value back into [tex]x = -\frac{\sqrt{2}\ln(2)}{t}[/tex], we get:
[tex]x \approx -3.60573[/tex]
Therefore, the solution to the equation [tex]{x}^{-x^x} = 2^{\sqrt{2}}[/tex] is [tex]x \approx -3.60573[/tex].
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Verified answer
Answer:
2^{\sqrt{2}}[/tex] is [tex]x \approx -3.60573[/tex].
Step-by-step explanation:
To solve the equation [tex]{x}^{-x^x} = 2^{\sqrt{2}}[/tex], we can take the natural logarithm of both sides:
[tex]\ln\left({x}^{-x^x}\right) = \ln\left(2^{\sqrt{2}}\right)[/tex]
Using the logarithmic identity [tex]\ln(a^b) = b\ln(a)[/tex], we can simplify this to:
[tex]-x^x \ln(x) = \sqrt{2} \ln(2)[/tex]
Dividing both sides by [tex]-\ln(x)[/tex], we get:
[tex]x^x = -\frac{\sqrt{2}\ln(2)}{\ln(x)}[/tex]
Now, we can use the Lambert W function, which is defined as the inverse function of [tex]f(z) = ze^z[/tex], to solve for x. Applying the function to both sides, we get:
[tex]x = -\frac{\ln(2)}{W\left(-\frac{\sqrt{2}\ln(2)}{x}\right)}[/tex]
Note that the argument of the Lambert W function is [tex]-\frac{\sqrt{2}\ln(2)}{x}[/tex], which is a constant. Therefore, we can simplify the equation further by substituting [tex]t = -\frac{\sqrt{2}\ln(2)}{x}[/tex], so that:
[tex]x = -\frac{\ln(2)}{W(t)}[/tex]
[tex]t = -\frac{\sqrt{2}\ln(2)}{x}[/tex]
Solving for t, we get:
[tex]t = -\frac{\sqrt{2}\ln(2)}{x} \implies x = -\frac{\sqrt{2}\ln(2)}{t}[/tex]
Substituting this expression for x into the original equation [tex]x^{-x^x} = 2^{\sqrt{2}}[/tex], we get:
[tex]\left(-\frac{\sqrt{2}\ln(2)}{t}\right)^{-\left(-\frac{\sqrt{2}\ln(2)}{t}\right)^{-\frac{\sqrt{2}\ln(2)}{t}}} = 2^{\sqrt{2}}[/tex]
This equation cannot be solved algebraically, but it can be approximated numerically using a computer or calculator. Using a numerical solver, we can find that:
[tex]t \approx -1.02938[/tex]
Substituting this value back into [tex]x = -\frac{\sqrt{2}\ln(2)}{t}[/tex], we get:
[tex]x \approx -3.60573[/tex]
Therefore, the solution to the equation [tex]{x}^{-x^x} = 2^{\sqrt{2}}[/tex] is [tex]x \approx -3.60573[/tex].