Let's solve the equations from the image.

4. Solve for $x$:

(a) $4^{2x-1} = 1024$

We can rewrite both sides using powers of 2: $4^{2x-1} = (2^2)^{2x-1} = 2^{4x - 2}$ Also, $1024 = 2^{10}$.

Now, we have: $2^{4x-2} = 2^{10}$

Equating the exponents: $4x - 2 = 10$

Solve for $x$: $4x = 12 \quad \Rightarrow \quad x = 3$

(b) $3^{2(9x - 2)} = 27^{x+1}$

Since $27 = 3^3$, we can rewrite the equation as: $3^{2(9x - 2)} = 3^{3(x+1)}$

Equating the exponents: $2(9x - 2) = 3(x + 1)$

Expand both sides: $18x - 4 = 3x + 3$

Solve for $x$: $18x - 3x = 3 + 4 \quad \Rightarrow \quad 15x = 7 \quad \Rightarrow \quad x = \frac{7}{15}$

(c) If $2^y = 24$, determine the value of $x$ and $z$ if $x = 2^{y - 2}$ and $z = 3(2^{y + 1})$.

First, solve for $y$: $2^y = 24$ Take the natural logarithm of both sides: $y \ln 2 = \ln 24 \quad \Rightarrow \quad y = \frac{\ln 24}{\ln 2}$ Using approximate logarithm values: $y = \frac{\ln 24}{\ln 2} \approx \frac{3.1781}{0.6931} \approx 4.585$

Now, solve for $x$: $x = 2^{y - 2} = 2^{4.585 - 2} = 2^{2.585} \approx 6.059$

Finally, solve for $z$: $z = 3(2^{y+1}) = 3(2^{4.585 + 1}) = 3(2^{5.585}) \approx 3 \times 48.09 = 144.27$

Do you have any questions or would you like further details?

Related Questions:

1. How can I simplify exponents when both sides have powers of the same base?
2. How do logarithmic properties help in solving exponential equations?
3. What is the general process for converting a non-power base to one with a common base?
4. Can we solve for other variables using natural logarithms for bases that aren't powers of 2 or 3?
5. How can I apply this method to solving more complex exponent problems?

Tip: When solving exponential equations, always try to express both sides with the same base for easier comparison of exponents.