A tricky arithmetic progression question

Question

In a question I have to prove that if $\log_l x, \log_m x, \log_n x$ are in AP where $x \neq 1$ and $x > 0$, prove that $$n^2=(l \cdot n)^{\log_l m}$$

My tries:

1. I first converted every term to natural logarithm so I got ln (x)/ln (l), ln (x)/ln (m), ln (x)/ln (n)
2. then I multiplied each term by (-1/ln (x)) because there is no x in result so I got ln (l),ln (m), ln (n) which are in AP
3. then I used three AP formulas A.M.=(a+b)/2 where a.m., a, b are arithmetic mean, a is first term, b is second term T base n =a+(n-1) d where T base n is nth term, d is common difference 2b=a+c, where a, b, c are 1st ,2nd and 3rd term respectively. But on using all these three formulas I am getting only one relation m^2=nl Please help how I should move forward to solve question

Answer 1

If $\log_a x, \log_b x, \log_c x$ are in AP, then we have $$ \log_a x + \log_c x = 2 \log_b x $$

Now, we express everything in base $c$ using the change of base formula: $$ \dfrac{\log_c x}{\log_c a} + \log_c x = \dfrac{2 \log_c x}{\log_c b} $$ After this, we divide both sides by $\log_c x$, then multiply both sides by $\log_c b$. $$ \begin{align*} \dfrac{1}{\log_c a} + 1 &= \dfrac{2}{\log_c b} \\ \dfrac{\log_c b}{\log_c a} + \log_c b &= 2 \end{align*} $$ Now, because $\dfrac{\log_c b}{\log_c a} = \log _a b $ by the change of base formula, we have $$ \begin{align*} \log_a b + \log_c b &= 2 \end{align*}$$

Taking $c$ to the power of both sides, we have the following: $$\begin{align*} c^{\log_a b + \log_c b} &= c^2 \\ c^{\log_a b}c^{\log_c b} &= c^2 \\ c^{\log_a b} \cdot b &= c^2 \\ \end{align*} $$

I think the new edit to the problem has parentheses wrong.

Answer 2

So $\frac{\ln x}{\ln l}, \frac{\ln x}{\ln m}, \frac{\ln x}{\ln n}$ are in AP, which means $$ \frac{2\ln x}{\ln m} = \frac{\ln x}{\ln l} + \frac{\ln x}{\ln n} $$ or equivalently $$ \frac{2}{\ln m} = \frac{1}{\ln l} + \frac{1}{\ln n} = \frac{\ln n + \ln l}{\ln l \ln n} $$ so $$ 2 \ln n = \frac{\ln m}{\ln l} \left( \ln n + \ln l \right). $$ Can you finish it?