1. Stating the problems: (a) Find the value of $\left( \frac{1}{4} \right)^{0.5}$. (b) Find the value of $(-8)^{\frac{2}{3}}$. (c) Calculate $2 \times 10^{100} - 2 \times 10^{98}$ and express the answer in standard form. 2. Solve (a): Recall that raising a number to the power $0.5$ is equivalent to taking its square root: $$\left( \frac{1}{4} \right)^{0.5} = \sqrt{\frac{1}{4}} = \frac{\sqrt{1}}{\sqrt{4}} = \frac{1}{2}$$ 3. Solve (b): Write $-8$ as $-8 = - (2^3)$. We use the property of exponents: $$(-8)^{\frac{2}{3}} = \left( -(2^3) \right)^{\frac{2}{3}} = \left(-1\right)^{\frac{2}{3}} \times \left(2^3\right)^{\frac{2}{3}}$$ Since $\left(-1\right)^{\frac{2}{3}} = \left((-1)^2\right)^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$. And by the power of a power rule: $$\left(2^3\right)^{\frac{2}{3}} = 2^{3 \times \frac{2}{3}} = 2^2 = 4$$ Therefore: $$(-8)^{\frac{2}{3}} = 1 \times 4 = 4$$ 4. Solve (c): Start with the expression: $$2 \times 10^{100} - 2 \times 10^{98}$$ Factor out $2 \times 10^{98}$: $$2 \times 10^{98} \left(10^2 - 1\right) = 2 \times 10^{98} \left(100 - 1\right) = 2 \times 10^{98} \times 99 = 198 \times 10^{98}$$ Express in standard form by converting $198$ into $1.98 \times 10^2$: $$198 \times 10^{98} = 1.98 \times 10^2 \times 10^{98} = 1.98 \times 10^{100}$$ Final answers: (a) $\frac{1}{2}$ (b) $4$ (c) $1.98 \times 10^{100}$