I came across a question in my textbook that I couldn't solve nor find the solution of:
If $b^2 = ac$, and $a, b, c$ are positive integers that are unnequal to each other and not equal to 1, prove that $$\frac{\log_a x}{\log_c x} = \frac{\log_a x - \log_b x}{\log_b x - \log_c x}$$
From the first statement, it is known that $2 \log b = \log a + \log c$, but evaluating it always results in an expression without any logarithm of x: $$\begin{align*} \frac{\log_a x}{log_c x} & = \frac{\frac{\log_b x}{\log_b a}}{\frac{\log_b x}{\log_b c}} \\ & = \frac{\log_b c}{\log_b a} \end{align*}$$ I didn't know how to continue from this.
Any help is appreciated!
$$\begin{array}{l} \displaystyle\frac{{{{\log }_a}x - {{\log }_b}x}}{{{{\log }_b}x - {{\log }_c}x}} = \frac{{{{\log }_a}x - {{\log }_a}x{{\log }_b}a}}{{{{\log }_c}x{{\log }_b}c - {{\log }_c}x}} = \frac{{{{\log }_a}x}}{{{{\log }_c}x}}.\frac{{1 - {{\log }_b}a}}{{{{\log }_b}c - 1}}\\ {b^2} = ac \Rightarrow {\log _b}{b^2} = {\log _b}a + {\log _b}c \Leftrightarrow {\log _b}a + {\log _b}c = 2\\ \displaystyle\Rightarrow {\log _b}a = 2 - {\log _b}c \Rightarrow \frac{{{{\log }_a}x - {{\log }_b}x}}{{{{\log }_b}x - {{\log }_c}x}} = \frac{{{{\log }_a}x}}{{{{\log }_c}x}}.\frac{{1 - 2 + {{\log }_b}c}}{{{{\log }_b}c - 1}} = \frac{{{{\log }_a}x}}{{{{\log }_c}x}}.\frac{{{{\log }_b}c - 1}}{{{{\log }_b}c - 1}}\\ \displaystyle= \frac{{{{\log }_a}x}}{{{{\log }_c}x}} \end{array}$$