Applied the properties of integer exponents to generate equivalent expressions to and parentheses seven minus power of three outside of the parentheses a power of 7×7 power of 20/6 with negative power two with only positive exponents then solve the expression provide the answer as an improper fraction

1 answer

To manipulate the expression you provided, we'll break it down step by step. It sounds like the expression has the following components:

  1. \((7^{-3})\)
  2. \(7^{7 \times 7^{20/6}}\)
  3. \((-2)\) as a negative exponent

Let's clarify the expression first. I will assume it looks something like this:

\[ (7^{-3}) \times (7^{7 \times 7^{20/6}}) \times (7^{-2}) \]

Step 1: Apply the properties of exponents

Using the properties of integer exponents, we can combine the powers of \(7\):

  • The property that states \(a^m \times a^n = a^{m+n}\).

Thus, we simplify:

\[ 7^{-3} \times 7^{7 \times 7^{20/6}} \times 7^{-2} = 7^{-3 - 2 + 7 \times 7^{20/6}} \]

Step 2: Evaluate the exponent

Now, simplify the exponent:

  1. The expression \(20/6\) simplifies to \(10/3\).

  2. Therefore \(7^{20/6} = 7^{10/3}\).

  3. Hence \(7 \times 7^{20/6} = 7 \times 7^{10/3} = 7^{1 + 10/3} = 7^{3/3 + 10/3} = 7^{13/3}\).

Now, substitute back into the expression:

\[ 7^{-3 - 2 + 7^{13/3}} = 7^{-5 + 7^{13/3}} \]

This simplifies to:

\[ 7^{7^{13/3} - 5} \]

Step 3: Convert any negative exponent

Since we want the answer only with positive exponents, we can express the entirety as:

\[ \frac{1}{7^{5 - 7^{13/3}}} \]

Final Expression

Given the above, the expression simplifies down to the proper form. We will convert this back to an improper fraction where needed.

\[ \frac{1}{7^{5 - \frac{7^{10}}{2}}} \]

Conclusion About Solving

To complete solving this expression will depend on knowing the specific numerical values obeying those ranges as both \(5\) and \(7^{10}\) will include substantial values:

Assuming numerical evaluations of primes were the task, rather than laying out a literal improper fraction, thus the final algebraic expression suffices unless further context or numerics to evaluate is applied.

If you want specific numerical values to be processed (like computing \(7^{10}\)), please let me know!