Question

If 4^2n-1 - 16^n-1 = 384. Then Find the value of n.

Answer 1

Hi...☺

Here is your answer...✌
=================================

${4}^{(2n - 1)} - {16}^{(n - 1)} = 384 \\ \\ 2 ^{2(2n - 1)} - {2}^{4(n - 1)} = 384 \\ \\ {2}^{(4n - 2)} - {2}^{(4n - 4)} = 384 \\ \\ {2}^{(4n - 2)} - {2}^{(4n - 2) - 2} = 384 \\ \\ {2}^{(4n - 2)} -\frac{ 2^{(4n - 2)}} {2^2} = 384 \\ \\ {2}^{(4n - 2)} (1 - \frac{1}{4} ) = 384 \\ \\ {2}^{(4n - 2)} \times \frac{3}{4} = 384 \\ \\ {2}^{(4n - 2)} = 384\times \frac{4}{3} \\ \\ {2}^{(4n - 2)} = 512 \\ \\ {2}^{(4n - 2)} = {2}^{9} \\ \\On\: equating \: the \: power \: of \: 2 \\ we \: get , \\ \\ 4n - 2 = 9 \\ \\ 4n = 11 \\ \\ n = \frac{11}{4}$

Answer 2

Hi,

Please see the attached file!


Thanks