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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A,B,C) Given $3^x = 4^{x-1}$.Taking $\log_3$ on both sides:$x = (x-1) \log_3 4$$x = (x-1) \cdot 2 \log_3 2$$x = 2x \log_3 2 - 2 \log_3 2$$2 \log_3 2

Vedclass · Accepted Answer

$(C) \frac{1}{1 - \log_4 3}$

Vedclass · Answer

(A,B,C) Given $3^x = 4^{x-1}$.Taking $\log_3$ on both sides:$x = (x-1) \log_3 4$$x = (x-1) \cdot 2 \log_3 2$$x = 2x \log_3 2 - 2 \log_3 2$$2 \log_3 2 = x(2 \log_3 2 - 1)$$x = \frac{2 \log_3 2}{2 \log_3 2 - 1}$ (Option $A$ is correct).Taking $\log_2$ on both sides:$x \log_2 3 = (x-1) \log_2 4$$x \log_2 3 = 2(x-1)$$x \log_2 3 = 2x - 2$$2 = x(2 - \log_2 3)$$x = \frac{2}{2 - \log_2 3}$ (Option $B$ is correct).Since $\log_2 3 = \log_4 3^2 = 2 \log_4 3$,we have:$x = \frac{2}{2 - 2 \log_4 3} = \frac{1}{1 - \log_4 3}$ (Option $C$ is correct).

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

Similar Questions

If $y = \log_x x$, then the value of $\left( \frac{dy}{dx} \right)^2$ at $x = e^2$ is

If $\frac{1}{2} \le \log_{0.1} x \le 2$,then:

The equation $x^{(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5}=3 \sqrt{3}$ has

If ${x^{\frac{3}{4}(\log_3 x)^2 + \log_3 x - \frac{5}{4}} = \sqrt{3}}$,then $x$ is:

If $\log _{e}\left(x^{2}-16\right) \leq \log _{e}(4 x-11)$, then

Vedclass Test Series

Exam Paper Generator

Online Exam Module