Question

Find the value of n if 2^2n-1=1/8^n-3​

Answer 1

$2^{2n-1} = \frac{1}{8^{n-3}}$

$\implies 2^{2n-1} = \frac{1}{(2^{3})^{n-3}}$

$\implies 2^{2n-1} = \frac{1}{2^{3(n-3)}}$

$\implies 2^{2n-1} = 2^{-3(n-3)}$

$\implies 2n-1 = -3(n-3)$

$\boxed{\pink { Since, If \:a^m = a^n \implies m = n }}$

$\implies 2n - 1 = -3n + 9$

$\implies 2n + 3n = 9 + 1$

$\implies 5n = 10$

$\implies n = \frac{10}{5}$

$\implies n = 2$

Therefore.,

$\red { Value \:of \:n } \green { = 2 }$

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