Question

(4^3 x 7^-2)^-5
_______________
7^9 x 4^-13.

Answer 1

To simplify the expression \((4^3 \times 7^{-2})^{-5} \div (7^9 \times 4^{-13})\), we first simplify the numerator and the denominator separately.

**Numerator:**

Start with the numerator:

\[
(4^3 \times 7^{-2})^{-5}
\]

Using the power of a product property \((a \times b)^n = a^n \times b^n\):

\[
= (4^3)^{-5} \times (7^{-2})^{-5}
\]

Now compute each part:

\[
(4^3)^{-5} = 4^{-15} \quad \text{(since } 3 \times -5 = -15\text{)}
\]

\[
(7^{-2})^{-5} = 7^{10} \quad \text{(since } -2 \times -5 = 10\text{)}
\]

So, the numerator becomes:

\[
4^{-15} \times 7^{10}
\]

**Denominator:**

Now simplify the denominator:

\[
7^9 \times 4^{-13}
\]

**Now put it all together:**

The entire expression is:

\[
\frac{4^{-15} \times 7^{10}}{7^9 \times 4^{-13}}
\]

Using the property of fractions \(\frac{a^m}{a^n} = a^{m-n}\):

Separate the bases:

- For \(4\):

\[
\frac{4^{-15}}{4^{-13}} = 4^{-15 - (-13)} = 4^{-15 + 13} = 4^{-2}
\]

- For \(7\):

\[
\frac{7^{10}}{7^9} = 7^{10 - 9} = 7^1 = 7
\]

So, combining these results, we have:

\[
4^{-2} \times 7 = 7 \times 4^{-2}
\]

Or in a more standard form:

\[
\frac{7}{4^2} = \frac{7}{16}
\]

Thus, the simplified expression is:

\[
\frac{7}{16}
\]