Solve aor \( x \) in the follong equations. \( x^{x}=8 \)

We wish to solve   xˣ = 8. A common method for equations where the variable appears both in the base and the exponent is to take logarithms. Here’s one way to proceed: 1. Write the equation as   xˣ = 8        (1) 2. Taking the natural logarithm of both sides gives   ln(xˣ) = ln(8). 3. Using the logarithm power rule (ln(aᵇ) = b·ln(a)), we have   x · ln(x) = ln(8). Since 8 = 2³, it follows that ln(8) = 3 ln2. Hence, we have   x · ln(x) = 3 ln2.      (2) 4. At this point it is useful to make the substitution   u = ln(x) ⟹ x = eᵘ. Then equation (2) becomes   eᵘ · u = 3 ln2. This equation is of the form   u · eᵘ = constant, which is exactly the type of equation solved by the Lambert W function. Recall that the Lambert W function, W(z), is defined as the function satisfying   W(z) e^(W(z)) = z. 5. Compare   u · eᵘ = 3 ln2    with  W(3 ln2) e^(W(3 ln2)) = 3 ln2. Thus, we can identify   u = W(3 ln2). 6. But u was defined as ln(x), so   ln(x) = W(3 ln2). Taking exponentials of both sides then gives   x = exp(W(3 ln2)). There is also an equivalent way to write the solution. Recall the property   exp(W(z)) = z/W(z). Thus, we may also write   x = (3 ln2) / W(3 ln2). This is the exact solution in closed form. To summarize, the solution for x in the equation xˣ = 8 is:   x = exp(W(3 ln2))     or equivalently,   x = (3 ln2) / W(3 ln2). Because the Lambert W function is not generally an elementary function, this answer is considered fully acceptable. Numerical evaluation will yield approximately x ≈ 2.39, which indeed satisfies xˣ ≈ 8.