Let's solve this exercise step by step. Here's a breakdown of each part.

Exercise 4

1. Simplify $A$ and $B$:

• Given:
  • $A = 5^{-3} \times (-10)^2 \times 2^{-3}$
  • $B = ((-3)^2)^4 \times \sqrt{81} \times \frac{1}{2^7}$

Solution for $A$:

1. $5^{-3} = \frac{1}{5^3} = \frac{1}{125}$.

2. $(-10)^2 = 100$.

3. $2^{-3} = \frac{1}{2^3} = \frac{1}{8}$.

   Therefore, $A = \frac{1}{125} \times 100 \times \frac{1}{8} = \frac{100}{1000} = \frac{1}{10} = 0.1$

Solution for $B$:

1. $(-3)^2 = 9$, so $((-3)^2)^4 = 9^4 = 6561$.

2. $\sqrt{81} = 9$.

3. $\frac{1}{2^7} = \frac{1}{128}$.

   Therefore, $B = 6561 \times 9 \times \frac{1}{128} = \frac{59049}{128}$

2. Given non-zero real numbers $a$ and $b$:

Expression: $C = \frac{(a^{-3} \times b)^5 \times (a^2)^{-9}}{(a^3 \times b^{-1})^4 \times b^{18}}$

Part (a): Simplify $C$

Let's simplify each part in the expression for $C$:

1. Numerator: $(a^{-3} \times b)^5 \times (a^2)^{-9}$

   • $(a^{-3} \times b)^5 = a^{-15} \times b^5$
   • $(a^2)^{-9} = a^{-18}$

   So, the numerator becomes: $a^{-15} \times b^5 \times a^{-18} = a^{-33} \times b^5$

2. Denominator: $(a^3 \times b^{-1})^4 \times b^{18}$

   • $(a^3 \times b^{-1})^4 = a^{12} \times b^{-4}$

   So, the denominator becomes: $a^{12} \times b^{-4} \times b^{18} = a^{12} \times b^{14}$

3. Combine the numerator and denominator: $C = \frac{a^{-33} \times b^5}{a^{12} \times b^{14}} = a^{-33 - 12} \times b^{5 - 14} = a^{-45} \times b^{-9} = \frac{1}{a^{45} b^9}$

   Thus, the simplified form of $C$ is: $C = \frac{1}{a^{45} b^9}$

Part (b): For $a = 10^{-3}$ and $b = 2 \times 10^{-2}$, find the scientific notation of $C$.

1. Substitute $a = 10^{-3}$ and $b = 2 \times 10^{-2}$ into $C$: $C = \frac{1}{(10^{-3})^{45} \times (2 \times 10^{-2})^9}$

2. Calculate each part:

   • $(10^{-3})^{45} = 10^{-135}$
   • $(2 \times 10^{-2})^9 = 2^9 \times (10^{-2})^9 = 512 \times 10^{-18} = 5.12 \times 10^{-16}$

   So, $C = \frac{1}{10^{-135} \times 5.12 \times 10^{-16}} = \frac{1}{5.12 \times 10^{-151}} = \frac{1}{5.12} \times 10^{151} \approx 0.1953 \times 10^{151} = 1.953 \times 10^{150}$

Would you like further details or have any questions?

Related Questions

1. What are the properties of exponents used in these simplifications?
2. How does scientific notation simplify handling extremely large or small numbers?
3. What is the general formula for simplifying expressions with negative exponents?
4. How can these simplifications be applied to real-world problems?
5. What are some common mistakes to avoid when simplifying expressions with multiple exponents?

Tip

When simplifying expressions with multiple terms in both the numerator and denominator, tackle each component separately, then combine them at the end for clarity.