Question

find the value of n if (3/4)^2n-5 = (27/64)^2-3n​

Answer 1

Answer:

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Answer 2

$Given \: \Big( \frac{3}{4}\Big)^{2n-5} = \Big( \frac{27}{64}\Big)^{2-3n}$

$\implies \Big( \frac{3}{4}\Big)^{2n-5} = \Big( \frac{3^{3}}{4^{3}}\Big)^{2-3n}$

$\implies \Big( \frac{3}{4}\Big)^{2n-5} = \Big( \frac{3}{4}\Big)^{3(2-3n)}$

$\boxed{ \pink{ \because \frac{x^{n}}{y^{n}} = \big( \frac{x}{y}\big)^{n} }}$

$\implies \Big( \frac{3}{4}\Big)^{2n-5} = \Big( \frac{3}{4}\Big)^{6 -9n)}$

$\implies 2n - 5 = 6 - 9n$

$\boxed{\blue{ If \: a^{m} = a^{n} \implies m = n }}$

$\implies 2n + 9n = 6 + 5$

$\implies 11n = 11$

$\implies n = \frac{11}{11}$

$\implies n = 1$

Therefore.,

$\red{ Value \: of\: n }\green { = 1 }$

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