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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:

  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) whih 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
Kartik Watwani
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    Do you have a picture/screenshot of the question that you could post, either there is a typo your type-up or I'm missing something! :) – Zain Patel Jun 24 '15 at 12:48
  • The best person to ask this is probably the one who asked you the question, since as it stands 'ln' does not mean anything just as you point out. – Ove Ahlman Jun 24 '15 at 12:51
  • A book is asking me this question. – Kartik Watwani Jun 24 '15 at 12:54
  • Presumably $l$ and $n$ are both numbers... – Najib Idrissi Jun 24 '15 at 12:54
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    @ChickenP I have changed your edit slightly because it seems that the while the OP seems to think $\ln$ should be natural log the way he wrote the question is consistent with the idea that $ln$ is just a product. In which case while the equality is strange and under specified it might make some sense given extra information. – DRF Jun 24 '15 at 12:56
  • OP can you give us more context of the question if you have it? Are there any extra equations? Or maybe some extra assumptions? – DRF Jun 24 '15 at 12:57
  • Given the picture (though sadly of very poor quality and with many of the interesting bits cut off) the correct interpretation is obviously that of product rather then $\ln$. The typography clearly shows slanted $ln$ but the usual roman $\log$. – DRF Jun 24 '15 at 13:00
  • I also feel now l and m are numbers – Kartik Watwani Jun 24 '15 at 13:00

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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.

Brian M. Scott
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