Question

solve this equation.​

Answer 1

n=1.

We'll follow the BODMAS rule to determine which operation shall be given priority.

Then we'll convert all bases in same format

Next we will add up the indices/powers of same bases to get single simplified term.

Finally we will compare both sides of the equation to get the value of n.

Answer 2

Answer:

$given \: that \: - \\ ( { \frac{3}{2} )}^{6} \div { (\frac{3}{2}) }^{(2n + 1)} \times {( \frac{3}{2} )}^{ - 3} = 1 \\ \\ to \: solve \: this \: using \: bodmas \: rule \: \\ = > (( { \frac{3 }{2} )}^{6} \div { (\frac{3}{2}) }^{(2n + 1)} ) \times ( { \frac{3}{2}) }^{ - 3} = 1 \\ we \: know \: that \: \frac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n} \\ \: \\ = > {( \frac{3}{2} )}^{(6 - 2n - 1)} \times { (\frac{3}{2}) }^{ - 3} = 1 \\ = > { (\frac{3}{2} )}^{(5 - 2n)} \times { (\frac{3}{2}) }^{ - 3} = 1 \: \\ we \: know \: that \: {a}^{m} \times {a}^{n} = {a}^{(m + n)} \\ \: = > {( \frac{3}{2}) }^{(5 - 2n - 3)} = 1 \\ \: = > { (\frac{3}{2} )}^{(2 - 2n)} = 1 \\ \: = > { (\frac{3}{2} }^{(2 - 2n)} = { (\frac{3}{2}) }^{0} \\ \: (since \: {a}^{0} = 1) \\ \: if \: bases \: are \: equal \: then \: exponents \: must \: be \: equal. \: \\ = > 2 - 2n = 0 \: \\ = > 2n = 2 \: \\ = > n = \frac{2}{2} \\ \: = > n = 1 \\ \: the \: value \: of \: n = 1$