Let's go through each question from the uploaded image and solve them one by one:

4.1

$\left( \frac{5}{13} \right)^0$ Anything raised to the power of zero is 1, so: $\left( \frac{5}{13} \right)^0 = 1$

4.2

$2^x \cdot 5^x$ We can combine the terms since both have the exponent $x$: $2^x \cdot 5^x = (2 \cdot 5)^x = 10^x$

4.3

$\frac{1}{2 + 2^{-1}}$ First, simplify $2^{-1} = \frac{1}{2}$: $2 + 2^{-1} = 2 + \frac{1}{2} = \frac{4 + 1}{2} = \frac{5}{2}$ So, the expression becomes: $\frac{1}{\frac{5}{2}} = \frac{2}{5}$

4.4

$\frac{2^{2n-1} \cdot 4^{n+1} \cdot 2}{4^n}$ Rewrite $4^{n+1}$ and $4^n$ in terms of base 2, since $4 = 2^2$: $4^{n+1} = (2^2)^{n+1} = 2^{2(n+1)} = 2^{2n+2}$ $4^n = (2^2)^n = 2^{2n}$ Substitute back into the expression: $\frac{2^{2n-1} \cdot 2^{2n+2} \cdot 2}{2^{2n}}$ Combine the exponents in the numerator: $= \frac{2^{(2n-1) + (2n+2) + 1}}{2^{2n}} = \frac{2^{4n+2}}{2^{2n}} = 2^{(4n+2) - 2n} = 2^{2n+2}$

4.5

$\frac{(a^3)^{n+1} \cdot (ab^2)^{-2n}}{(ab)^n \cdot b^{-5n}}$ Rewrite each part using power rules:

1. $(a^3)^{n+1} = a^{3(n+1)} = a^{3n+3}$
2. $(ab^2)^{-2n} = a^{-2n} \cdot b^{-4n}$
3. $(ab)^n = a^n \cdot b^n$

Substitute and simplify: $= \frac{a^{3n+3} \cdot a^{-2n} \cdot b^{-4n}}{a^n \cdot b^n \cdot b^{-5n}}$ Combine $a$ and $b$ exponents separately: $= a^{(3n+3 - 2n - n)} \cdot b^{(-4n - n + 5n)} = a^3 \cdot b^0 = a^3$

4.6

$\frac{2 \cdot 3^x}{3^{x+2} - 3^x}$ Factor $3^x$ out of the denominator: $= \frac{2 \cdot 3^x}{3^x (3^2 - 1)} = \frac{2 \cdot 3^x}{3^x \cdot 8} = \frac{2}{8} = \frac{1}{4}$

4.7

Solve for $x$: $2^x = 64$ Rewrite $64$ as a power of $2$: $64 = 2^6$ Thus: $2^x = 2^6 \Rightarrow x = 6$

4.8

Solve for $x$: $3^x = \frac{1}{81}$ Rewrite $\frac{1}{81}$ as $3^{-4}$: $3^x = 3^{-4} \Rightarrow x = -4$

Would you like a deeper explanation for any of these steps?

Additional Practice Questions:

1. Simplify $\left(\frac{3}{4}\right)^{-2}$.
2. Solve for $x$ in $5^{2x} = 125$.
3. Rewrite $(x^2 y^{-3})^2$ with positive exponents only.
4. Evaluate $\frac{16^{-1/2}}{4^{-1}}$.
5. Solve $7^{x+1} = 343$ for $x$.

Tip:

When simplifying expressions with exponents, remember that terms with the same base can be combined by adding or subtracting their exponents based on multiplication or division.