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Identify the base of the multiplication.

Chemistry

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Answer

(57)5\left(\frac{5}{7}\right)^5

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Step-by-step solution

8 steps

Here are the solutions for problems 6 and 7:

6. Express in power form: 57×57×57×57×57\frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7} \times \frac{5}{7}

  1. Identify the base of the multiplication.

    The base is 57\frac{5}{7}.

  2. Count how many times the base is multiplied by itself.

    The base 57\frac{5}{7} is multiplied by itself 5 times.

  3. Write the expression in power form.

    (57)5\left(\frac{5}{7}\right)^5

    The expression in power form is:

    (57)5\boxed{\left(\frac{5}{7}\right)^5}

    7. Evaluate: (311)−1×(311)2×(311)3×(311)−4\left(\frac{3}{11}\right)^{-1} \times \left(\frac{3}{11}\right)^2 \times \left(\frac{3}{11}\right)^3 \times \left(\frac{3}{11}\right)^{-4}

  4. Identify the common base.

    The common base is 311\frac{3}{11}.

  5. Apply the exponent rule am×an=am+na^m \times a^n = a^{m+n} to combine the terms.

    Add all the exponents:

    −1+2+3+(−4)-1 + 2 + 3 + (-4)
  6. Calculate the sum of the exponents.

    −1+2+3−4=1+3−4=4−4=0-1 + 2 + 3 - 4 = 1 + 3 - 4 = 4 - 4 = 0
  7. Substitute the sum of the exponents back into the expression.

    (311)0\left(\frac{3}{11}\right)^0
  8. Evaluate the expression.

    Any non-zero number raised to the power of 0 is 1.

    (311)0=1\left(\frac{3}{11}\right)^0 = 1

    The evaluated expression is:

    1\boxed{1}
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