Determine under what conditions the following equalities are equivalent:

$$
\frac{\log _{a} t}{\log _{c} t}=\frac{\log _{a} t-\log _{b} t}{\log _{b} t-\log _{c} t} \quad \text { and } \quad b=\sqrt{a c} \text {. }
$$
Both sides of both equalities are defined if the following conditions are met:

$$
a>0, \quad b>0, \quad c>0, \quad t>0, \quad a \neq 1, \quad b \neq 1, \quad c \neq 1, \quad t \neq 1 \text { and } b \neq c
$$

We claim that under these conditions, the two equalities are equivalent. We show this by transforming the first equality into the second through equivalent transformations.

Let's switch to base $t$ logarithms in the first equality, using the fact that if $t>0, t \neq 1, x>0$ and $x \neq 1$, then $\log _{x} t=\frac{1}{\log _{t} x}$. The left side then becomes $\frac{\log _{t} c}{\log _{t} a}$, and the right side becomes

$$
\frac{\frac{1}{\log _{t} a}-\frac{1}{\log _{t} b}}{\frac{1}{\log _{t} b}-\frac{1}{\log _{t} c}}=\frac{\log _{t} c}{\log _{t} a} \cdot \frac{\log _{t} b-\log _{t} a}{\log _{t} c-\log _{t} b}
$$

Given our conditions, $\frac{\log _{t} c}{\log _{t} a} \neq 0$, so simplifying by this term, we get the equivalent equality

$$
1=\frac{\log _{t} b-\log _{t} a}{\log _{t} c-\log _{t} b}
$$

Since $c \neq b$, the denominator on the right side of (1) is not zero, so (1) holds if and only if

$$
\log _{t} c-\log _{t} b=\log _{t} b-\log _{t} a
$$

We know that $a, b$, and $c$ are all positive, so (2) is true if and only if

$$
\log _{t} \frac{c}{b}=\log _{t} \frac{b}{a}
$$

The logarithmic function is bijective, so for $t>0, t \neq 1$, (3) is equivalent to

$$
\frac{c}{b}=\frac{b}{a}
$$

Finally, (4) is clearly true if and only if

$$
a c=b^{2}
$$

which is

$$
\sqrt{a c}=b, \text { since } a, b \text { and } c \text { are positive numbers. }
$$

We have thus shown that if the first equality is meaningful, it can indeed be transformed into the second through equivalent transformations, thus completing the solution.