Question

Please solve this and ignore if you don't know the answer or I will report.​

Answer 1

Required Answer:-

Given:

$\sf\leadsto \dfrac{ {a}^{2n - 3} \times ( {a}^{2})^{n + 1} }{ ({a}^{4})^{ - 3} } = ({a}^{3})^{3} \div ({a}^{6})^{ - 3}$

To find:

• The value of n.

Solution:

We have,

$\sf \implies\dfrac{ {a}^{2n - 3} \times ( {a}^{2})^{n + 1} }{ ({a}^{4})^{ - 3} } = ({a}^{3})^{3} \div ({a}^{6})^{ - 3}$

$\sf \implies\dfrac{ {a}^{2n - 3} \times{a}^{2(n + 1)}}{{a}^{ - 12}} = ({a}^{9}) \div ({a}^{ - 18}) \: \: \blue{\bigg( ({a}^{m})^{n} = {a}^{mn} \bigg) }$

$\sf \implies\dfrac{ {a}^{2n - 3 + 2n + 2} }{{a}^{ - 12}} = ({a}^{9}) \div ({a}^{ - 18}) \: \: \blue{\bigg( {a}^{m}\times {a}^{n} = {a}^{m + n} \bigg) }$

$\sf \implies\dfrac{ {a}^{4n - 1} }{{a}^{ - 12}} = ({a}^{9}) \div ({a}^{ - 18})$

$\sf \implies\dfrac{ {a}^{4n - 1} }{{a}^{ - 12}} = ({a}^{9}) \times \dfrac{1}{ ({a}^{ - 18}) }$

$\sf \implies{a}^{4n - 1} \times {a}^{12} = ({a}^{9} \times {a}^{18} ) \: \: \: \blue{ \bigg( {a}^{n} = \dfrac{1}{ {a}^{ - n} } \bigg)}$

$\sf \implies{a}^{4n - 1 + 12} = {a}^{9 + 18} \: \: \: \blue{ \bigg( {a}^{m} \times {a}^{n} = {a}^{m + n} \bigg)}$

$\sf \implies{a}^{4n + 11} = {a}^{27}$

Comparing base, we get,

$\sf \implies 4n + 11 = 27$

$\sf \implies 4n = 27 - 11$

$\sf \implies 4n = 16$

$\sf \implies n =4$

★ Hence, the value of n is 4.

Answer:

• The value of n is 4.