In a question I have to prove that if log (base l of x), log (base m of x), log (base n of x) are in AP where x doesnt equals 1 and x is positive, prove that n^2=(l*n)^(log base l of m)> My tries:
- I first converted every term to natural logarithm so I got ln (x)/ln (l), ln (x)/ln (m), ln (x)/ln (n)
- 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) whih are in AP
- 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
Your second step isn’t right: when you multiply by $\frac{-1}{\ln x}$, you get $\frac1{\ln\ell}$, $\frac1{\ln m}$, and $\frac1{\ln n}$ in arithmetic progression, not $\ln\ell$, $\ln m$, and $\ln n$.
I would approach it a bit differently. We want to show that $n^2=(\ell n)^{\log_\ell m}$, or, equivalently, that $n^2=mn^{\log_\ell m}$. If we take logs base $\ell$, we see that this in turn is equivalent to
$$2\log_\ell n=\log_\ell m(1+\log_\ell n)\;.\tag{0}$$
This suggests that we should investigate $\log_\ell m$ and $\log_\ell n$.
Let $d$ be the common difference of the original arithmetic progression, and let $a=\log_\ell x$; then
$$a+d=\log_mx=\frac{\log_\ell x}{\log_\ell m}=\frac{a}{\log_\ell m}\;,$$
so
$$\log_\ell m=\frac{a}{a+d}\;.\tag{1}$$
A similar calculation shows that
$$\log_\ell n=\frac{a}{a+2d}\;.\tag{2}$$
Since, as you say, there is no $x$ in the desired equality, we need to get rid of $a$. If we let $b=d/a$, we can rewrite $(1)$ and $(2)$ as
$$\log_\ell m=\frac1{1+b}$$
and
$$\log_\ell n=\frac1{1+2b}\;.$$
The target equality $(0)$ can then be rewritten in the form
$$\frac2{1+2b}=\frac1{1+b}\left(1+\frac1{1+2b}\right)\;,$$
which is easily verified.