In the following question I fail to understand why the A option is correct.
I understand that D is wrong, and that B and C are correct, but why is A correct?
If $3^x=4^{x-1}$, then $x $cannot be equal to:
A. $2\log_32$
B. $\frac{2}{2-\log_23}$
C. $\frac{1}{1-log_43}$
D. $\frac{2\log_23}{2\log_23-1}$
By taking $\log_3$ on both sides, we get $x=2(x-1)\log_32$.
After that we get B and C as correct after making modifications but how do we know that A is correct?
$$x=2\log_3{2} \\ 3^{2\log_3{2}}=4^{2\log_3{2}-1} \\ 2^2=\frac{4^{2\log_3{2}}}{4} \\ 16 \neq 16^{\log_3{2}}$$ Therefore the expression cannot equal A.