If 3 x = 4 x - 1 , then x =

If $3^{x} = 4^{x - 1}$ , then x =

$\frac{{2 log}_{3} 2}{{2 log}_{3} 2 - 1}$
$\frac{2}{2 - {log}_{2} 3}$
$\frac{1}{1 - {log}_{4} 3}$
$\frac{{2 log}_{2} 3}{{2 log}_{2} 3 - 1}$

Solution

3^x = 4^x^{- 1}
x = (x - 1) log₃ 4
x ( 1 - 2 log₃ 2) = -2log₃ 2

x = \frac{2 {log}_{3} 2}{2 {log}_{3} 2 - 1}

x log₂ 3 = (x - 1).2
x (log₂ 3 - 2) = -2

x = \frac{2}{2 {- log}_{2} 3}

x = \frac{1}{1 - \frac{1}{2} {log}_{2} 3} = \frac{1}{1 - {log}_{4} 3}

Asked in: JEE Advanced 2013 (Paper 2)

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