If \(a^b = b^a\), then what is \(\frac{a \times (\frac{a}{b}) ^{ \frac{a}{b} } }{ (a)^{ \frac{a}{b} } }\) equal to?
Understanding the Problem
We are given an equation relating two variables, \(a^b = b^a\), and asked to find the value of a specific mathematical expression involving these variables:
\(E = \frac{a \times \left(\frac{a}{b}\right)^{ \frac{a}{b} } }{ a^{ \frac{a}{b} } }\)Our goal is to simplify this expression using the given condition.
Simplifying the Expression
Let's break down the simplification process step-by-step:
- Rewrite the expression using the properties of exponents. Specifically, we can rewrite the division as multiplication by a negative exponent: \(E = a \times \left(\frac{a}{b}\right)^{ \frac{a}{b} } \times a^{ - \frac{a}{b} }\)
- Group terms with the same base (\(a\) and \(b\)). Recall that \(\left(\frac{a}{b}\right)^{ \frac{a}{b} } = \frac{a^{ \frac{a}{b} }}{b^{ \frac{a}{b} }}\). Substituting this in the original expression gives: \(E = \frac{a \times \frac{a^{ \frac{a}{b} }}{b^{ \frac{a}{b} }} }{ a^{ \frac{a}{b} } }\) Now, simplify the fraction: \(E = a \times \frac{a^{ \frac{a}{b} }}{b^{ \frac{a}{b} }} \times \frac{1}{a^{ \frac{a}{b} }}\)
- Cancel out common factors. The term \(a^{ \frac{a}{b} }\) appears in both the numerator and the denominator: \(E = a \times \frac{1}{b^{ \frac{a}{b} }}\) \(E = \frac{a}{ b^{ \frac{a}{b} } }\)
Using the Condition \(a^b = b^a\)
Now we need to use the given condition \(a^b = b^a\) to simplify the expression \(E = \frac{a}{ b^{ \frac{a}{b} } }\).
Let's manipulate the condition:
- Start with \(a^b = b^a\).
- Raise both sides to the power of \(\frac{1}{b}\): \((a^b)^{\frac{1}{b}} = (b^a)^{\frac{1}{b}}\)
- Using the exponent rule \((x^m)^n = x^{mn}\), we get: \(a^{b \times \frac{1}{b}} = b^{a \times \frac{1}{b}}\) \(a^1 = b^{\frac{a}{b}}\) \(a = b^{\frac{a}{b}}\)
This gives us a relationship between \(a\) and \(b^{\frac{a}{b}}\).
Final Calculation
Substitute the result \(a = b^{\frac{a}{b}}\) back into our simplified expression for \(E\):
\(E = \frac{a}{ b^{ \frac{a}{b} } }\) \(E = \frac{a}{a}\) \(E = 1\)Therefore, the value of the given expression is 1.
Similar Questions
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If \(x^4 + y^4 = 14x^2 y^2\), then consider the following :
I. \(\log_{10}(x^2 + y^2)\) = \(\log_{10} x + \log_{10} y + 2\log_{10} 2\)
II. \(\log_{10}(x^2-y^2)=\log_{10} x + \log_{10} y\) +\(\log_{10}2+0.5\log_{10} 3\)
Which of the above is/are correct?
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II. The logarithm of a positive number less than unity is negative.
Which of the statements given above is/are correct?
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If \(\log_ba = p\), \(\log_dc = 2p\) and \(\log_fe = 3p\), then what is \((ace)^{\frac{1}{p}}\) equal to ?
