For n and p > 0, the expression (p^2n1/2)^8 √p^5n^4 equivalent to p^18n^6√p? Justify your answer.
Yes, the expression (p^2n1/2)^8√p^5n^4 is equivalent to p^18n^6√p.
To simplify the expression, we can first use the exponent rule that states (a^b)^c = a^(bc). Applying this rule to the first term in the expression, we have:
(p^2n1/2)^8 = p^(2n1/2 * 8) = p^(8n)
Next, we can simplify the second term by using the exponent rules for multiplication. We have:
√p^5n^4 = (p^5n^4)^(1/2) = p^(5/2)n^(4/2) = p^(5/2)n^2
Now we can simplify the entire expression by multiplying the two terms together and combining the exponents of p and n. We have:
(p^(8n))*(p^(5/2)n^2) = p^(8n+5/2)n^2
This means that the expression is equivalent to p^(8n+5/2)n^2√p. To make this match the form of the given equivalent expression, we can multiply the whole expression by p^(5/2) and divide it by p^(5/2). We get:
(p^(8n+5/2)n^2√p) * (p^(5/2)/p^(5/2)) = p^(8n+5/2+5/2)n^2√p
Now we have p^(8n+5)n^2√p, which is equivalent to p^18n^6√p. Therefore, we have shown that (p^2n1/2)^8√p^5n^4 is equivalent to p^18n^6√p.
To determine if the expression (p^2n1/2)^8 √p^5n^4 is equivalent to p^18n^6√p, we need to simplify both expressions and compare them.
Let's simplify the first expression:
(p^2n1/2)^8 √p^5n^4
First, we use the exponent rule that states (a^b)^c = a^(b * c):
[(p^2n1/2)^8] * √p^5n^4
Next, we apply the exponent rule to both terms inside the bracket and simplify:
(p^(2*8) * (n1/2)^8) * √p^5n^4
= p^16 * (n^(1/2))^8 * √p^5n^4
Simplifying further, we use the property that (a * b)^c = a^c * b^c:
= p^16 * (n^((1/2)*8)) * √(p^5 * n^4)
= p^16 * n^4 * √(p^5 * n^4)
Now, let's simplify the second expression:
p^18n^6√p
We can't simplify this expression further because we cannot combine the exponent of n with the square root of p.
Comparing the two expressions, we can see that they are not equivalent. The first expression simplifies to p^16 * n^4 * √(p^5 * n^4), while the second expression is p^18 * n^6 * √p.
Therefore, (p^2n1/2)^8 √p^5n^4 is not equivalent to p^18n^6√p.