<s>[INST] $\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243$ , find the value of $x$. $\log_{3}\frac{16^{x-3}+14}{4^{x-3}+2}=243=3^5$
$\Rightarrow \frac{16^{x-3}+14}{4^{x-3}+2}= {3^3}^5=3^{243}$
I am not able to proceed from here. Please help !!!
The options given for this problem are :-

*

*$x$ is a rational number.

*$x$ is a natural number.

*$x$ is an even natural number.

*$x$ is a rational number less than $0$.

Thanks in advance !!!
 [/INST] Let $x$  starts from $-\infty$ and jumps only on integers waits at $0$ and again starts journey towards $\infty$
Let,
$$f(x) = \frac {16^{x-3} + 14}{4^{x -3} + 2}$$
$$\implies \log_3  \frac {16^{x-3} + 14}{4^{x -3} + 2} = L(x) = \log_3(f(x))$$

*

*$x\in (-\infty, 0)$
$$\lim_{x\to-\infty}f(x) = 7 \text{ so also the value of }L(x\to-\infty) \to \text{some constant number}$$
Let me interpret this as the derivative of $L(x)$ should tend to $0$ as the value of $x\to-\infty$ or $\color{blue}{\text{Should we say this as the value of $L(x)$ is not changing as values of $x$ does?}} $

*

*$x=0$
$f(0) \approx 7$ which is far enough to not to expect $L(0) = 243$


*$x\in (0, \infty)$
Lets write a function: $\lambda(x)$ = $\text{last digit function giving you the last digit of $x$}$
$$\begin{align}
\\ \lambda(3^2)=9
\\ \lambda(3^3)=7
\\ \lambda(3^4)=1
\\ \lambda(3^5)=3
\\
\\
\\ \implies \lambda(3^{243}) = 7
\end{align}$$
similarly
$$\lambda(L(i)) = \frac {\lambda(16^{i-3})+14}{\lambda(4^{i-3})+ 2} = \frac {\text{something ending with zero}}{\text{something ending with 6 or 8}} \ne \text{something ending with 7}$$
$x \notin \mathbb{Z}$
</s>