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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.AAE.13

13. For what x>0 does x^(x^x) = (x^x)^x? Give reasons for your answer.

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Start by writing the given equation clearly: \(x^{x^{x}} = (x^{x})^{x}\) for \(x > 0\).
Rewrite the right-hand side using the power of a power rule: \((x^{x})^{x} = x^{x \cdot x} = x^{x^{2}}\).
Now the equation becomes \(x^{x^{x}} = x^{x^{2}}\). Since the bases are the same and \(x > 0\), set the exponents equal: \(x^{x} = x^{2}\).
Rewrite the equation \(x^{x} = x^{2}\) by taking the natural logarithm of both sides: \(\ln(x^{x}) = \ln(x^{2})\).
Use the logarithm power rule to simplify: \(x \ln(x) = 2 \ln(x)\). Since \(x > 0\), consider cases where \(\ln(x) \neq 0\) and solve for \(x\).

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Properties of Exponents

Understanding how to manipulate and simplify expressions with exponents is crucial. Key rules include (a^b)^c = a^(bc) and a^b * a^c = a^(b+c). These properties help rewrite and compare expressions like x^(x^x) and (x^x)^x by expressing them with a common base or exponent form.
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Exponentiation with Variable Exponents

Exponentiation where the exponent itself is a function of the variable, such as x^x or x^(x^x), requires careful handling. Recognizing the hierarchy of operations and how to interpret nested exponents is essential to correctly simplify and analyze the expressions.
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Equation Solving and Domain Considerations

Solving the equation x^(x^x) = (x^x)^x involves setting the expressions equal and simplifying to find x > 0. Considering the domain ensures the expressions are defined, and applying logarithms or exponent rules helps isolate x and determine all valid solutions.
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