Question

help me immediately please​

Answer 1

Answer:

n=4

Step-by-step explanation:

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: Answer:-}}}}}$

◉GIVEN:-

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

◉ TO FIND:-

• value of n.

◉SOLUTION:-

$: \longrightarrow \: \frac{ {3}^{2n} \times {3}^{2} \times {3}^{n} - {3}^{3n} }{ {3}^{3m} \times {2}^{3} } = \frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{(2n + 2 + n)} - {(3}^{3) {}^{n} } }{ {3}^{15} \times {2}^{3} } = \frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{(3n + 2)} - ({3}^{ {3)}^{n} } }{ {3}^{15} \times {2}^{3} } = \frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{3n} \times {3}^{2} - {3}^{3n} }{ {3}^{15} \times {2}^{3} } = \frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{3n} \times ( {3}^{2} - 1)}{ {3}^{15} \times {2}^{3} } = \frac{1}{27}$

$: \longrightarrow \: = \frac{ {3}^{3n} \times ( 9 - 1)}{ {3}^{15} \times {2}^{3} }\frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{3n} \times ( 8)}{ {3}^{15} \times {2}^{3} } = \frac{1}{27}$

As 8 can also be written as 2^3.

$: \longrightarrow \: \frac{ {3}^{3n} \times ( \cancel{{2}^{3} )}}{ {3}^{15} \times \cancel{{2}^{3} }} = \frac{1}{27}$

$: \longrightarrow \: \frac{ {3}^{3n}}{ {3}^{15}} = \frac{1}{27}$

Taking 3^15 to the numerator

$: \longrightarrow \: {3}^{3n - 15} = \frac{1}{27}$

$: \longrightarrow \: {3}^{3n - 15} = \frac{1}{ {3}^{3} }$

$: \longrightarrow \: {3}^{3n - 15} = {3}^{ - 3}$

Comparing the powers:-

$: \longrightarrow\: 3n - 15 = - 3$

$: \longrightarrow\: 3n = - 3 + 15$

$: \longrightarrow\: 3n = 12$

$: \longrightarrow\: n = \frac{12}{3}$

${ \boxed{ \underline{ \bf{ \blue{n = 4}}}}}$

HENCE,

the value of n is 4.