Question

simplify each of the following, leaving your answer in index notation where appropriate
(a) $3^{7} \times 8^{7}$
(b) $(5b^{4})^{3}$
(c) $(-2c^{2}d^{5})^{5}$
(d) $\dfrac{(h^{4}k^{2})^{2} \times (-5hk^{5})^{3}}{(5h^{3}k)^{3}}$

Explanation:

Part (a)

Step 1: Recall the exponent rule \(a^n \times b^n=(a\times b)^n\)

We have \(3^{7} \times 8^{7}\), using the rule \((a\times b)^n=a^n\times b^n\) in reverse, we get \((3\times8)^{7}\)

Step 2: Calculate \(3\times8\)

\(3\times8 = 24\), so \((3\times8)^{7}=24^{7}\)

Step 1: Recall the exponent rule \((ab)^n=a^n\times b^n\)

For \((5b^{4})^{3}\), we apply the rule \((ab)^n=a^n\times b^n\) where \(a = 5\) and \(b=b^{4}\) and \(n = 3\)

Step 2: Apply the rule to each part

\(5^{3}\times(b^{4})^{3}\), we know that \(5^{3}=125\) and \((b^{m})^{n}=b^{m\times n}\), so \((b^{4})^{3}=b^{4\times3}=b^{12}\)

Step 3: Combine the results

\(125\times b^{12}=125b^{12}\)

Step 1: Recall the exponent rule \((ab)^n=a^n\times b^n\)

For \((-2c^{2}d^{5})^{5}\), we apply the rule \((ab)^n=a^n\times b^n\) to each factor \(-2\), \(c^{2}\) and \(d^{5}\)

Step 2: Apply the rule to each part

\((-2)^{5}\times(c^{2})^{5}\times(d^{5})^{5}\), we know that \((-2)^{5}=-32\), \((c^{m})^{n}=c^{m\times n}\) so \((c^{2})^{5}=c^{2\times5}=c^{10}\) and \((d^{5})^{5}=d^{5\times5}=d^{25}\)

Step 3: Combine the results

\(-32\times c^{10}\times d^{25}=-32c^{10}d^{25}\)

Answer:

\(24^{7}\)

Part (b)