Find the value of the product

$\begin{matrix}4^{X_1}=5 \\ (4^{X_1})^{X_2}=5^{X_2}=6 \\ (4^{X_1X_2})^{X_3}=6^{X_3}=7 \\ \vdots \\ \vdots \\ \vdots \\ (4^{X_1X_2X_3\cdots \cdots X_{122}})^{X_{123}}=127 \\ (4^{X_1X_2X_3\cdots \cdots X_{123}})^{X_{124}}=128 \end{matrix}$

Therefore, $2^{2(X_1X_2X_3\cdots \cdots X_{123}X_{124})}=2^7$

Since, the bases are equal...

So, $2X_1X_2X_3\cdots \cdots X_{123}X_{124}=7$

$\implies \displaystyle \prod^{124}_{i=1}X_i=X_1X_2X_3\cdots \cdots X_{123}X_{124}=\dfrac{7}{2}=\boxed{3.5}$

Note that $\small (k+3)^{X_k} = (k+4) \implies X_k = \dfrac{\log(k+4)} {\log(k+3)}, k\in\mathbb Z^+$ $P = \prod_{k=1}^{123} X_k = \prod_{k=1}^{123} \dfrac{\log\,(k+4)}{\log\,(k+3)} \\P = \dfrac{1}{\log(4)} \times\log(128) \times \underbrace{\prod_{k=1}^{123}\dfrac{\log(k+4)}{\log(k+4)}}_{\text{Telescoping product}}= \dfrac{7}{2} = \boxed{3.5}$