Question

remember, you are not handing in scratch paper, but work that shows your thinking and supports your responses. 1. simplify: \\(\left(2b^{12}\
ight)^4 \left(4b^8\
ight)^{-\frac{1}{2}} = \frac{16^{48}}{2^4}\\)

Explanation:

Step1: Simplify \((2b^{12})^4\)

Using the power of a product rule \((ab)^n = a^n b^n\) and power of a power rule \((a^m)^n = a^{mn}\), we have:
\((2b^{12})^4 = 2^4 \cdot (b^{12})^4 = 16b^{48}\)

Step2: Simplify \((4b^8)^{-\frac{1}{2}}\)

First, rewrite \(4\) as \(2^2\), so \(4b^8 = 2^2b^8\). Then use the power of a product rule and negative exponent rule \(a^{-n}=\frac{1}{a^n}\):
\((2^2b^8)^{-\frac{1}{2}}=(2^2)^{-\frac{1}{2}}\cdot (b^8)^{-\frac{1}{2}} = 2^{-1}\cdot b^{-4}=\frac{1}{2b^4}\)

Step3: Multiply the two simplified expressions

Multiply \(16b^{48}\) and \(\frac{1}{2b^4}\):
\(16b^{48}\cdot\frac{1}{2b^4}=\frac{16}{2}\cdot\frac{b^{48}}{b^4}\)
Using the quotient rule for exponents \(\frac{a^m}{a^n}=a^{m - n}\) and simplifying \(\frac{16}{2}=8\), we get:
\(8b^{48 - 4}=8b^{44}\)

Wait, the given answer \(\frac{16^{48}}{2^4}\) seems incorrect. Let's re - check the original problem's intended simplification. Maybe there was a mis - input in the problem or the given answer. But following the correct exponent rules:

If we consider the original problem as \(\frac{(2b^{12})^4}{(4b^8)^{\frac{1}{2}}}\) (maybe a sign error in the exponent). Then:

\((2b^{12})^4 = 16b^{48}\), \((4b^8)^{\frac{1}{2}}=(2^2b^8)^{\frac{1}{2}} = 2b^4\)

Then \(\frac{16b^{48}}{2b^4}=8b^{44}\)

If we strictly follow the given expression \((2b^{12})^4(4b^8)^{-\frac{1}{2}}\), the correct simplification is \(8b^{44}\). But if we assume the given answer's numerator and denominator are about the coefficients (ignoring the \(b\) terms which might be a mistake in the problem), \(\frac{16^48}{2^4}\) is incorrect. However, if we consider the coefficient part of \((2b^{12})^4 = 16\) and \((4b^8)^{-\frac{1}{2}}=\frac{1}{2}\) (ignoring the \(b\) terms which is wrong, but just for the coefficient part), \(16\times\frac{1}{2} = 8\), not \(\frac{16^{48}}{2^4}\).

But if we consider the problem as a pure coefficient problem (ignoring the \(b\) terms which is non - standard for an algebraic expression with \(b\)):

\((2)^4\times(4)^{-\frac{1}{2}}=16\times\frac{1}{2}=8\), and \(\frac{16}{2}=8\), \(\frac{16^1}{2^1}\) (not \(16^{48}\)). There must be a mistake in the given answer or the problem statement.

But if we follow the given answer's structure (maybe a mis - written problem where the exponents are on the coefficients only):

\((2)^4 = 16\), \((4)^{-\frac{1}{2}}=\frac{1}{2}\), then \(16\times\frac{1}{2}=\frac{16}{2}=\frac{2^4}{2}=\ 2^{3}=8\). But \(\frac{16^{48}}{2^4}\) is \(2^{192}\div2^4 = 2^{188}\), which is wrong.

Assuming that the problem was supposed to be \(\frac{(2)^{48}}{(4)^{\frac{1}{2}}}\) (a mis - input of exponents):

\(4^{\frac{1}{2}} = 2\), so \(\frac{2^{48}}{2}=2^{47}\), still not matching \(\frac{16^{48}}{2^4}\).

16 is \(2^4\), so \(16^{48}=(2^4)^{48}=2^{192}\), \(2^4 = 16\), \(\frac{2^{192}}{2^4}=2^{188}\), which is not a simple simplification.

It seems there is a mistake in the problem or the given answer. But if we proceed with the correct exponent rules for the algebraic expression \((2b^{12})^4(4b^8)^{-\frac{1}{2}}\), the answer is \(8b^{44}\).

Answer:

The correct simplification of \((2b^{12})^4(4b^8)^{-\frac{1}{2}}\) is \(8b^{44}\). The given answer \(\frac{16^{48}}{2^4}\) is incorrect.