Log-Power Inversion Identity

We have :
$$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$$
we get $$3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$$
we get $$\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$$
we get $$4+\log_4\left(x-1\right)\ =\ 3$$
$$\log_4\left(x-1\right)\ =\ -1$$
x-1 = 4^-1 
x = $$\frac{1}{4}+1=\frac{5}{4}$$
4x = 5

$$\log _{10}{2} = \log _{10}{10} - \log _{ 10}{5}$$
So, $$\log _{ 10}{2} = 1 - \log _{10}{5}$$
Hence, the first option is correct.

As, $$(1+2+3)=(1*2*3)=6$$
So, $$\log _{10} (1+2+3) = \log _{10} (1*2*3)$$
Hence, the third option is correct.

However, $$2 * \log_{10}{5} = \log_{10}{25} > 1$$

If $$Y<1, \log _{10}{Y}$$ is negative and the equation has no solution. So, $$Y>1$$ and $$Y^{Y}$$ = 10.
The function $$Y^{Y}$$ is a monotonically increasing function and $$2^{2} < 10$$ and $$3^{3}>10$$.
So, there exists exactly one Y between 2 and 3 such that $$Y = \log _{10}{Y}$$

$$\log _{10}{Y}$$