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

(A) $\frac{2 \log _{3} 2}{2 \log _{3} 2-1}$; (B) $\frac{2}{2-\log _{2} 3}$; (C) $\frac{1}{1-\log _{4} 3}$

(D) $\frac{2 \log _{2} 3}{2 \log _{2} 3-1}$

Basic Mathematics — JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 Question with Solution

3^{x} = 4^{x - 1}

, then

x =

Answer: (A), (B), (C)

Starting with

3^{x} = 4^{x - 1}

, take

lo g_{3}

of both sides:

x = (x - 1) lo g_{3} 4 = (x - 1) \cdot 2 lo g_{3} 2

x = 2 x lo g_{3} 2 - 2 lo g_{3} 2

x (1 - 2 lo g_{3} 2) = - 2 lo g_{3} 2

x = \frac{2 l o g _{3} 2}{2 l o g _{3} 2 - 1}

This is option (A). For option (B), using the change of base formula:

lo g_{3} 2 = \frac{1}{l o g _{2} 3}

Substituting:

x = \frac{2/ l o g _{2} 3}{2/ l o g _{2} 3 - 1} = \frac{2}{2 - l o g _{2} 3}

For option (C):

lo g_{4} 3 = \frac{l o g _{2} 3}{2}

, so

\frac{1}{1 - l o g _{4} 3} = \frac{2}{2 - l o g _{2} 3}

Option (D) would give a different value, so it is incorrect.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options