Question

plz answer my question ​

Answer 1

Solution:

$\rm : \implies \bigg( \dfrac{3}{2} \bigg) {}^{5} \times \bigg( \dfrac{3}{2} \bigg) {}^{4} = \bigg( \dfrac{3}{2} \bigg) {}^{2n + 1}$

using exponential law

$\rm\bigg( \dfrac{3}{2} \bigg) {}^{5 + 4} =\bigg( \dfrac{3}{2} \bigg) {}^{2n + 1}$

$\rm\bigg( \dfrac{3}{2} \bigg) {}^{9} = \bigg( \dfrac{3}{2} \bigg) {}^{2n + 1}$

Now we can write

$\rm \: 9 = 2n + 1$

$\rm \: 2n = 9 - 1$

$\rm \: 2n = 8$

$\rm \: n = \dfrac{8}{2}$

$\rm \: n \: = 4$

So value of n = 4

•Some exponential law

$\to \rm \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\to \rm \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\to \rm \: a {}^{ - 1} = \dfrac{1}{a}$

$\to \rm {a}^{0} = 1$

$\rm \to(a {}^{m} ) {}^{n} = {a}^{mn}$

$\to \rm \sqrt[2]{a} = {a}^{ \frac{1}{2} }$

Answer 2

PLEASE SEE THE ABOVE ATTACHMENT.