१. सर्व संख्यांचा पाया 3 करू:
- $81 = 3^4 \implies 81^x = 3^{4x}$
- $27 = 3^3 \implies 27^3 = (3^3)^3 = 3^9$
- दिलेले पद: $3^2$ आणि उजवी बाजू: $3 = 3^1$
२. समीकरणात पाया 3 ठेवून मांडणी:
$$(3^{4x} \div 3^9) \times 3^2 = 3^1$$
३. घातांकांचे नियम वापरू:
$$3^{4x - 9 + 2} = 3^1$$
$$3^{4x - 7} = 3^1$$
४. पाया समान असल्याने घातांक समान करू:
$$4x - 7 = 1$$
$$4x = 8$$
$$x = \frac{8}{4} = \mathbf{2}$$
५. पडताळा:
$$\left(\frac{81^2}{27^3}\right) \times 9 = \frac{6561}{19683} \times 9 = \frac{1}{3} \times 9 = 3\text{ (अगदी बरोबर)}$$
1. Express all numbers in terms of base 3:
- $81 = 3^4 \implies 81^x = (3^4)^x = 3^{4x}$
- $27 = 3^3 \implies 27^3 = (3^3)^3 = 3^9$
- $3^2 = 3^2$
- $3 = 3^1$
2. Rewrite the original equation with base 3:
$$\left(3^{4x} \div 3^9\right) \times 3^2 = 3^1$$
3. Apply the laws of indices ($a^m \div a^n = a^{m-n}$ and $a^m \times a^n = a^{m+n}$):
$$3^{4x - 9} \times 3^2 = 3^1$$
$$3^{4x - 9 + 2} = 3^1$$
$$3^{4x - 7} = 3^1$$
4. Equate the exponents of the equal bases:
$$4x - 7 = 1$$
$$4x = 1 + 7 = 8$$
$$x = \frac{8}{4} = \mathbf{2}$$
5. Verification:
$$\left(81^2 \div 27^3\right) \times 3^2 = \left(6561 \div 19683\right) \times 9 = \frac{1}{3} \times 9 = 3\text{ (Exact match!)}$$
$(27^3) = (3^3)^3 = 3^9$ होते, $3^6$ नाही. घातांकाचा घातांक असल्यास घातांकांचा गुणाकार होतो.
Remember that $(27)^3 = (3^3)^3 = 3^9$, not $3^6$. Multiplying powers $(3 \times 3 = 9)$ is required by $(a^m)^n = a^{mn}$.
$$4x - 7 = 1 \implies 4x = 8 \implies x = \mathbf{2}$$!
$$3^{4x - 9 + 2} = 3^1 \implies 4x - 7 = 1 \implies 4x = 8 \implies x = \mathbf{2}$$!
१. 3: डावी बाजू 243 येते.
२. 2: $4(2) - 7 = 1 \implies 3^1 = 3$ (अचूक उत्तर).
३. 1: डावी बाजू 1/27 येते.
४. 10: चुकीचा पर्याय.
1. 3: Gives $3^{4(3)-7} = 3^5 = 243 \neq 3$.
2. 2: Gives $3^{4(2)-7} = 3^1 = 3$, verified!
3. 1: Gives $3^{4(1)-7} = 3^{-3} = 1/27 \neq 3$.
4. 10: Far too large exponent.