Question

{NO SPAMMING} only for genius @spammers stay away
answer should be explained and correct
​

Answer 1

Step-by-step explanation:

i hope it may help you ....

the value of n = 4

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• $\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{(3^3)^5 \times 2^3} = \dfrac{1}{27}$

To Find:

• Value of n

Solution:

$\dfrac{9^n \times 3^2 \times 3^n - (27)^n}{(3^3)^5 \times 2^3} = \dfrac{1}{27} \\\\\\\implies \dfrac{(3^2)^n \times 3^2 \times 3^n - (3^3)^n}{(3^3)^5 \times 2^3} = \dfrac{1}{3^3} \\\\\\\implies \dfrac{3^{2n} \times 3^2 \times 3^n - 3^{3n}}{3^{15} \times 2^3} = 3^{-3} \\\\\\\implies \dfrac{3^{2n+2+n} - 3^{3n}}{3^{15} \times 2^3} = 3^{-3} \\\\\\\implies \dfrac{3^{3n+2} - 3^{3n}}{3^{15} \times 2^3} = 3^{-3} \\\\\\\implies \dfrac{3^{3n} \times 3^2 - 3^{3n}}{3^{15} \times 2^3} = 3^{-3} \\\\\\\implies \dfrac{(3^{3n})\times (3^2 - 1)}{3^{15} \times 2^3} = 3^{-3} \:\:\:\:(\text{Taking 3^{3n} as common}) \\\\\\\implies \dfrac{3^{3n}\times 8}{3^{15} \times 2^3} = 3^{-3} \\\\\\\implies \dfrac{3^{3n}\times 2^3}{3^{15} \times 2^3} = 3^{-3} \\\\\implies 3^{3n-15} \times 2^{3-3} = 3^{-3} \\\\\implies 3^{3n-15} \times 2^0 = 3^{-3} \\\\\implies 3^{3n-15} = 3^{-3}\\\\\text{As bases are same, we compare the powers} \\\\\implies 3n - 15 = -3 \\\\\implies 3n = 12 \\\\\bf{\implies n = 4}$

Formula Used:

• $a^m \times a^n= a^{m+n}$
• $\dfrac{1}{a^n} = a^{-n}$
• $(a^m)^n = a^{mn}$
• $\dfrac{a^m}{a^n} = a^{m-n}$

Learn More:

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$

Hope it helps!!