Question

PLEASE GIVE THE ANSWER WITH THE SOLUTION...

Answer 1

$Answer \: \\ \\ Given \: Question \: Is \: \: \\ \\ log_{3}(4 \times 3 {}^{x} - 1 ) = 2x + 1 \\ \\ log_{3}(4 \times 3 {}^{x} - 1 ) = log_{3}(3 {}^{(2x + 1)} ) \\ \\ 4 \times 3 {}^{x} - 1 = 3 {}^{(2x + 1)} \\ \\ 4 \times 3 {}^{x} - 1= 3 {}^{2x} \times 3 {}^{1} \\ \\ \frac{3 {}^{x} }{3 {}^{2x} \times 3 {}^{1} } = \frac{1}{3 {}^{2x} \times 3 {}^{1} } \\ \\ 3 {}^{( - x - 1)} = 3 {}^{ - (2x + 1)} \\ \\ now \: compare \: powers \: of \: 3 \: we \: have \\ \\ ( - x - 1) = - (2x + 1) \\ \\ - x - 1 = - 2x - 1 \\ \\ - x - 1 + 2x + 1 = 0 \\ \\ x = 0 \\ \\ therefore \: \: x = 0 \\ \\ NOTE \: \\ \\ log_{y}(y) = 1 \\ \\ log(x {}^{n} ) = n log(x) \\ \\ \frac{1}{x {}^{n} } = x {}^{ - n}$