Simplifying Exponential Expressions A Step-by-Step Guide To Solving Complex Math Problems

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At the heart of this mathematical exploration lies the simplification of a complex exponential expression. The expression, meticulously crafted with nested exponents and roots, challenges us to unravel its inherent value. Let's break down the components and apply the fundamental rules of exponents to navigate through the intricacies of the problem.

Breaking Down the Expression

The given expression is:

${\left[\frac{9^{n+1/4} \cdot \sqrt{3} \cdot 3^n}{\sqrt[3]{3^{-n}}}\right]^{\frac{1}{n}}}$

Our initial step involves expressing all terms with the same base. We know that 9 is 3 squared, and roots can be represented as fractional exponents. Rewriting the expression, we get:

${\left[\frac{(3^2)^{n+1/4} \cdot 3^{1/2} \cdot 3^n}{(3^{-n})^{1/3}}\right]^{\frac{1}{n}}}$

Applying the power of a power rule, which states that (am)n=amn(a^m)^n = a^{mn}, we can further simplify the expression:

${\left[\frac{3^{2(n+1/4)} \cdot 3^{1/2} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Distributing the exponent in the numerator, we have:

${\left[\frac{3^{2n+1/2} \cdot 3^{1/2} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Combining Terms with the Same Base

Now, we can combine the terms in the numerator by adding the exponents, using the rule am⋅an=am+na^m \cdot a^n = a^{m+n}:

${\left[\frac{3^{2n+1/2 + 1/2 + n}}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Simplifying the exponents in the numerator, we get:

${\left[\frac{3^{3n+1}}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Next, we divide terms with the same base by subtracting the exponents, using the rule aman=am−n\frac{a^m}{a^n} = a^{m-n}:

${\left[3^{(3n+1)-(-n/3)}\right]^{\frac{1}{n}}}$

Simplifying the exponent, we have:

${\left[3^{3n+1+n/3}\right]^{\frac{1}{n}}}$

Combining the terms with 'n' in the exponent:

${\left[3^{\frac{10n}{3}+1}\right]^{\frac{1}{n}}}$

Final Simplification

Finally, we apply the power of a power rule again:

${3^{(\frac{10n}{3}+1)\frac{1}{n}}}$

Distributing the exponent 1n\frac{1}{n}, we get:

${3^{\frac{10}{3} + \frac{1}{n}}}$

However, there seems to be a slight discrepancy in the simplification process. Let's re-evaluate the steps to pinpoint the error. Upon closer inspection, the error lies in the final distribution of the exponent. The correct distribution should be:

${3^{(\frac{10n}{3}+1)\frac{1}{n}} = 3^{\frac{10n}{3n} + \frac{1}{n}} = 3^{\frac{10}{3} + \frac{1}{n}}}$

This is where the mistake occurred. The correct simplification after distributing 1n\frac{1}{n} should be:

${3^{\frac{10n}{3n} + \frac{1}{n}} = 3^{\frac{10}{3} + \frac{1}{n}}}$

But this is not leading us to a clear single numerical answer as in the options (1, 3, 9, 27). Let's backtrack and check each step meticulously to find the exact point of divergence.

Revisiting the step where we combined exponents after dividing, we had:

${\left[3^{(3n+1)-(-n/3)}\right]^{\frac{1}{n}}}$

Which simplifies to:

${\left[3^{3n+1+n/3}\right]^{\frac{1}{n}}}$

Combining the 'n' terms gives us:

${\left[3^{\frac{10n}{3}+1}\right]^{\frac{1}{n}}}$

Now, applying the power of a power rule:

${3^{(\frac{10n}{3}+1)\frac{1}{n}}}$

Distributing 1n\frac{1}{n}:

${3^{\frac{10n}{3} \cdot \frac{1}{n} + 1 \cdot \frac{1}{n}}}$

${3^{\frac{10}{3} + \frac{1}{n}}}$

Aha! The error is not in the algebraic manipulation but in the interpretation. We are looking for a value independent of 'n'. Let's go back to the original expression and try a different approach, focusing on factoring out terms.

A More Direct Approach

Starting from:

${\left[\frac{9^{n+1/4} \cdot \sqrt{3} \cdot 3^n}{\sqrt[3]{3^{-n}}}\right]^{\frac{1}{n}}}$

Rewrite as:

${\left[\frac{(3^2)^{n+1/4} \cdot 3^{1/2} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

${\left[\frac{3^{2n+1/2} \cdot 3^{1/2} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Combine terms in the numerator:

${\left[\frac{3^{2n+1/2+1/2} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

${\left[\frac{3^{2n+1} \cdot 3^n}{3^{-n/3}}\right]^{\frac{1}{n}}}$

${\left[\frac{3^{3n+1}}{3^{-n/3}}\right]^{\frac{1}{n}}}$

Now, divide by subtracting exponents:

${\left[3^{3n+1-(-n/3)}\right]^{\frac{1}{n}}}$

${\left[3^{3n+1+n/3}\right]^{\frac{1}{n}}}$

${\left[3^{\frac{10n}{3}+1}\right]^{\frac{1}{n}}}$

Apply the outer exponent:

${3^{(\frac{10n}{3}+1)\frac{1}{n}}}$

${3^{\frac{10n}{3n} + \frac{1}{n}}}$

${3^{\frac{10}{3} + \frac{1}{n}}}$

Still, we arrive at the same impasse. The expression contains 'n', and we need a numerical value. Let's consider if there's a misunderstanding of the question or a typo. Given the multiple-choice options, the expression should simplify to a constant. The most likely scenario is that the question intends for the expression to simplify such that the 'n' terms cancel out. Let's revisit the most crucial steps and see if we can force a cancellation by correctly interpreting the rules of exponents.

The Key Insight: Forcing Cancellation

Let’s rewind to the step where we have:

${\left[3^{\frac{10n}{3}+1}\right]^{\frac{1}{n}}}$

If the intended answer is a constant, the term 1n\frac{1}{n} must somehow interact to eliminate 'n'. The exponent rule we're using is (am)n=amn(a^m)^n = a^{mn}. Applying this, we get:

${3^{(\frac{10n}{3}+1) \cdot \frac{1}{n}}}$

${3^{\frac{10n}{3} \cdot \frac{1}{n} + 1 \cdot \frac{1}{n}}}$

${3^{\frac{10}{3} + \frac{1}{n}}}$

We are still stuck with the 1n\frac{1}{n} term. However, if we assume there was a slight oversight in the original question and the exponent outside the bracket was intended to perfectly cancel out the 'n' term, we need to rethink our approach. Perhaps the question meant something slightly different.

Let’s consider a hypothetical scenario where the original expression was designed such that after simplification, the power of 3 is a constant. If that’s the case, the most crucial part to examine is the exponent after combining all the terms inside the bracket.

We had:

${\left[3^{\frac{10n}{3}+1}\right]^{\frac{1}{n}}}$

The exponent inside the bracket is 10n3+1\frac{10n}{3} + 1. If the 1n\frac{1}{n} outside the bracket is meant to simplify this to a constant, the 'n' in the denominator should ideally cancel out the 'n' in the numerator. The only way this happens perfectly is if the entire exponent inside the bracket was a multiple of 'n'.

Let's hypothesize that the correct simplification inside the bracket should have led to something like 33n3^{3n}. If that were the case, then applying the 1n\frac{1}{n} would give us:

${(3^{3n})^{\frac{1}{n}} = 3^{3n \cdot \frac{1}{n}} = 3^3 = 27}$

This gives us one of the options: 27. However, this is a significant leap of assumption and implies there might be an error in the original question or a missing piece of information.

Given the constraints and the multiple-choice options, the most likely correct answer, derived by assuming a perfect cancellation of 'n' terms, is 27. But this comes with the strong caveat that we are making an assumption about the question's intent rather than a direct derivation.

Conclusion (with a Caveat)

Based on our analysis and the need to fit one of the multiple-choice options, the most plausible answer is (d) 27. However, it's crucial to acknowledge that this conclusion relies on an assumption that the question was designed to have a perfect cancellation of 'n' terms, which our direct algebraic simplification did not yield. There might be a subtle error in the original question or a missing piece of information that would lead to a more rigorous solution.

Simplify the expression: [9n+1/4⋅3⋅3n3−n3]1n\left[\frac{9^{n+1/4} \cdot \sqrt{3} \cdot 3^n}{\sqrt[3]{3^{-n}}}\right]^{\frac{1}{n}} and find its value.

Simplifying Exponential Expressions A Step-by-Step Guide to Solving Complex Math Problems