Question

2log(x+1)-log(x²-1)=log2​

Answer 1

$HEYA \: \\ \\ 2 log(x + 1) - log(x {}^{2} - 1 ) = log(2) \\ \\ log(x + 1) {}^{2} - log(x {}^{2} - 1 ) = log(2) \\ becoz \: \: \: n log(y) = log(y {}^{n} ) \\ \\ log((x + 1) {}^{2} \div (x {}^{2} - 1) ) = log(2) \\ becoz \: log(m) - log(n) = log(m \div n) \\ \\ (x + 1) {}^{2} = 2(x {}^{2} - 1) \\ \\ x {}^{2} - 2x {}^{2} + 1 + 2 + 2x = 0 \\ \\ - x {}^{2} + 2x + 3 = 0 \\ \\ x {}^{2} - 2x - 3 = 0 \\ \\ x {}^{2} - 3x + 1x - 3 = 0 \\ \\ x(x - 3) + 1(x - 3) = 0 \\ \\ (x + 1) = 0 \: \: or \: \: (x - 3) = 0 \\ \\ x = - 1 \: \: or \: \: x = 3 \\ \\ x = - 1 \: \: will \: be \: rejected \: becoz \: \\ it \: is \: not \: satisfying \: the \: original \: \\ equation \: \\ \\ so \: x = 3 \: is \: only \: solution \: for \: this \: equation.$