Question

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

• 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. $\frac{2{\log }_{2}3}{2{\log }_{2}3-1}$

Answer 1

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

The correct options are
A $\frac{2{\log }_{3}2}{2{\log }_{3}2-1}$
B $\frac{1}{1-{\log }_{4}3}$
C $\frac{2}{2-{\log }_{2}3}$

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

Taking $\log$ on both sides with a base $2$

$x{\log }_2\left(3\right)=(x-1){\log }_2\left(4\right)$.

$x{\log }_2\left(3\right)=x{\log }_2\left(4\right)-{\log }_2\left(4\right)$

${\log }_2\left(4\right)=x\left[{\log }_2\right(4)-{\log }_2(3\left)\right]$

$x=\frac{{\log }_2\left(4\right)}{{\log }_2\left(4\right)-{\log }_2\left(3\right)}$

$x=\frac{{\log }_2(2{)}^2}{{\log }_2(2{)}^2-{\log }_2(3)}=\frac{2}{2-{\log }_2\left(3\right)}...\left(1\right)(\because {\log }_a{a}^x=x)$

Now using, ${\log }_{a}b=\frac{\log b}{\log a}$

$x=\frac{2}{2-\frac{\log \left(3\right)}{\log \left(2\right)}}=\frac{2\log \left(2\right)}{2\log \left(2\right)-\log \left(3\right)}$ ..... (Taking L.C.M.)

Dividing by $\log \left(3\right)$ to numerator and denominator.

$x=\frac{2\frac{\log \left(2\right)}{\log \left(3\right)}}{2\frac{\log \left(2\right)}{\log \left(3\right)}-1}=\frac{2{\log }_3\left(2\right)}{2{\log }_3\left(2\right)-1}$

$x=\frac{2}{2-{\log }_2\left(3\right)}$ ......(from 1)

$x=\frac{2}{2-\frac{\log \left(3\right)}{\log \left(2\right)}}=\frac{1}{1-\frac{\log \left(3\right)}{2\log \left(2\right)}}$ (Dividing numerator and denominator by 2).

$x=\frac{1}{1-\frac{\log \left(3\right)}{\log \left(4\right)}}=\frac{1}{1-{\log }_4\left(3\right)}$