Let's solve each problem step by step.

(a)
$3^3 \cdot 9^6 \cdot 27 \div 81^2 = x$ We rewrite each term in terms of base $3$:

• $9 = 3^2 \Rightarrow 9^6 = (3^2)^6 = 3^{12}$
• $27 = 3^3$
• $81 = 3^4 \Rightarrow 81^2 = (3^4)^2 = 3^8$

Now substitute: $3^3 \cdot 3^{12} \cdot 3^3 \div 3^8 = x$ Using the rule $a^m \cdot a^n = a^{m+n}$: $3^{3 + 12 + 3} \div 3^8 = 3^{18} \div 3^8 = 3^{18 - 8} = 3^{10}$ Thus, $x = 3^{10} = 59049$.

(b)
$4^{(2x - 5)} = 32$ Express $32$ as a power of $4$: $32 = 4^{\frac{5}{2}}$ Since the bases are equal, equate the exponents: $2x - 5 = \frac{5}{2}$ Solve for $x$: $2x = \frac{5}{2} + 5 = \frac{5}{2} + \frac{10}{2} = \frac{15}{2}$ $x = \frac{15}{4} = 3.75$

(c)
$\log (x + 50) = 2.5$ Rewrite in exponential form: $x + 50 = 10^{2.5} = 10^{5/2} = \sqrt{10^5} = \sqrt{100000} = 316.23$ Solve for $x$: $x = 316.23 - 50 = 266.23$

(d)
$\log_3 (2x - 3) = 3$ Rewrite in exponential form: $2x - 3 = 3^3 = 27$ Solve for $x$: $2x = 27 + 3 = 30 \quad \Rightarrow \quad x = \frac{30}{2} = 15$

Summary of Results:

• (a) $x = 59049$
• (b) $x = 3.75$
• (c) $x = 266.23$
• (d) $x = 15$

Do you have any questions or want more details about any step?

Related Questions:

1. How do logarithm rules apply when changing bases?
2. What is the general method for rewriting exponents in terms of a common base?
3. Can you explain how to solve exponential equations with fractions as exponents?
4. What are other ways to express logarithmic equations in exponential form?
5. How do calculators handle fractional powers and logs with decimals?

Tip:
When solving equations with exponents, try rewriting terms with the same base to simplify calculations.