Question

no irrelevant answers please.
Answer it fast​

Answer 1

$\big(\frac{9}{4} \big)^{x+1} \div \big(\frac{2}{3} \big)^{3x-1}=\big(\frac{3}{2} \big)^{11}\\ \\ \Rightarrow \big(\frac{3}{2} \big)^{2(x+1)} \div \big(\frac{3}{2} \big)^{-1(3x-1)}=\big(\frac{3}{2} \big)^{11}\\ \\ \Rightarrow \big(\frac{3}{2} \big)^{2x+2} \div \big(\frac{3}{2} \big)^{-3x+1}=\big(\frac{3}{2} \big)^{11}\\ \\ \Rightarrow \big(\frac{3}{2} \big)^{2x+2-(-3x+1)} =\big(\frac{3}{2} \big)^{11}$

$\Rightarrow \big(\frac{3}{2} \big)^{2x+2+3x-1} = \big(\frac{3}{2} \big)^{11}\\ \\ \Rightarrow \big(\frac{3}{2} \big)^{5x+1} = \big(\frac{3}{2} \big)^{11} \\ \\ \text{comapring\ the\ powers} \\ \\ \Rightarrow 5x+1 = 11\\ \\ \Rightarrow 5x = 11-1 = 10 \\ \\ \Rightarrow x = \frac{10}{5}=2$

Answer 2

$\tt {( \frac{3}{2} )}^{2(x + 1)} \times{( \frac{3}{2} ) }^{ - 3x + 1} = {(\frac{3}{2} )}^{11}$

$\tt \: \: ⠀ \color{cyan}{☢} \: \: \red\lgroup{ \blue{ \: after \: \: \: cancelling \: \: \: ( \frac{3}{2} )} \red \rgroup}$

$\tt2x + 2 - ( - 3x +1) = 11$

$\tt2x + 2 + 3x - 1 = 11$

$\tt5x + 1 = 11$

$\tt5x = 11 - 1 = 10$

$\tt x = \frac{10}{5} = 2$

$\red{ \colorbox{pink}{x = 2}}$

the value of x is 2