a,b, c, are in geometric progression and log_b(a), log_c(b),log_a(c) are in arithmetic progression

Dhamnekar Winod

Active member
If a,b, c, are in G.P and $\log_ba, \log_cb,\log_ac$ are in A.P. I want to find the common difference of A.P.

After doing some computations, I stuck here. $\frac{2(\log a+\log r)}{\log a+2\log r}=\frac{2(\log a)^2+3\log r\log a +2(\log r)^2}{(\log a)^2+\log r\log a}$

How to proceed further? (a= 1st term in G.P. r= common ratio of G.P.)
Answer is $\frac32$. I don't understand how it is computed. If any member knowing further computations to arrive at the answer, may reply.

Last edited:

Olinguito

Well-known member
Hi Dhamnekar Winod .

I got the answer $-\dfrac5{12}$ or $0$ (the latter being the trivial case $a=b=c$). Either the given answer is wrong or (which IMHO is the more likely) you made a typo copying the question.

Assuming you copied the question correctly, however, this is my solution.

If $a$,$b$,$c$ are in GP, then

$b\ =\ \sqrt{ac}$ (assuming $a,b,c$ are all positive, which appears to be what the question is doing)​

$\implies\ \log_ab\ =\ \dfrac{1+\log_ac}2$.

If $\log_ab$, $\log_cb$, $\log_ac$ are in AP, then
$$\log_cb\ =\ \frac{\log_ab}{\log_ac}\ =\ \frac{\log_ab+\log_ac}2$$
and substituing for $\log_ab$,
$$\frac{\frac{1+\log_ac}2}{\log_ac}\ =\ \frac{\frac{1+\log_ac}2+\log_ac}2$$
which becomes
$$3\left(\log_ac\right)^2-\log_ac-2\ =\ \left(3\log_ac+2\right)\left(\log_ac-1\right)\ =\ 0$$

$\implies\ \log_ac=-\dfrac23\ \text{or}\ 1$.

In the former case, $\log_ab=\dfrac16$ and $\log_cb=-\dfrac14$ – and you can check that
$$\frac16,\ -\frac14,\ -\frac23$$
is an AP with common difference $-\dfrac5{12}$.

Last edited:

Dhamnekar Winod

Active member
Hi Dhamnekar Winod .

I got the answer $-\dfrac5{12}$ or $0$ (the latter being the trivial case $a=b=c$). Either the given answer is wrong or (which IMHO is the more likely) you made a typo copying the question.

Assuming you copied the question correctly, however, this is my solution.

If $a$,$b$,$c$ are in GP, then
$b\ =\ \sqrt{ac}$ (assuming $a,b,c$ are all positive, which appears to be what the question is doing)​

$\implies\ \log_ab\ =\ \dfrac{1+\log_ac}2$.

If $\log_ab$, $\log_cb$, $\log_ac$ are in AP, then
$$\log_cb\ =\ \frac{\log_ab}{\log_ac}\ =\ \frac{\log_ab+\log_ac}2$$
and substituing for $\log_ab$,
$$\frac{\frac{1+\log_ac}2}{\log_ac}\ =\ \frac{\frac{1+\log_ac}2+\log_ac}2$$
which becomes
$$3\left(\log_ac\right)^2-\log_ac-2\ =\ \left(3\log_ac+2\right)\left(\log_ac-1\right)\ =\ 0$$

$\implies\ \log_ac=-\dfrac23\ \text{or}\ 1$.

In the former case, $\log_ab=\dfrac16$ and $\log_cb=-\dfrac14$ – and you can check that
$$\frac16,\ -\frac14,\ -\frac23$$
is an AP with common difference $-\dfrac5{12}$.

Hello,
Your work is great. But i am sorry to say that the question posted was wrong. Please read $\log_ba$ instead of $\log_ab$.

Last edited:

Olinguito

Well-known member
Please read $\log_ba$ instead of $\log_ab$.
Ah, sorry it was my fault. It was I who copied your question incorrectly.

Now $a$, $b$, $c$ form a GP $\implies$ $b^2=ac$ $\implies$
$$2\log_cb\ =\ \log_ca+1.$$
If $\log_ba$, $\log_cb$, $\log_ac$ form an AP, then
$$2\log_cb\ =\ \log_ba+\log_ac$$
(note: $x$, $y$, $z$ form an AP $\implies$ $2y=x+z$); i.e.
$$2\log_cb\ =\ \frac{\log_ca}{\log_cb}+\frac1{\log_ca}$$
converting to base $c$.

However, substituting for $\log_ca$ or $\log_cb$ (working omitted) would give cubic equations simplifying to $\log_ca=\log_cb=1$, i.e. $a=b=c$, i.e.
$$\log_ba,\log_cb,\log_ac$$
is just $1,1,1$.

So I still don’t think the common difference is $\frac32$; it’s $0$ (i.e. a constant AP).

Dhamnekar Winod

Active member
Ah, sorry it was my fault. It was I who copied your question incorrectly.

Now $a$, $b$, $c$ form a GP $\implies$ $b^2=ac$ $\implies$
$$2\log_cb\ =\ \log_ca+1.$$
If $\log_ba$, $\log_cb$, $\log_ac$ form an AP, then
$$2\log_cb\ =\ \log_ba+\log_ac$$
(note: $x$, $y$, $z$ form an AP $\implies$ $2y=x+z$); i.e.
$$2\log_cb\ =\ \frac{\log_ca}{\log_cb}+\frac1{\log_ca}$$
converting to base $c$.

However, substituting for $\log_ca$ or $\log_cb$ (working omitted) would give cubic equations simplifying to $\log_ca=\log_cb=1$, i.e. $a=b=c$, i.e.
$$\log_ba,\log_cb,\log_ac$$
is just $1,1,1$.

So I still don’t think the common difference is $\frac32$; it’s $0$ (i.e. a constant AP).
Hello,

After some computations, I got $$4(\log_cb)^3-2(\log_cb)^2=2\log_ca\log_cb-\log_ca+\log_cb$$ How did you solve this?

Olinguito

Well-known member
I got $$4(\log_cb)^3-2(\log_cb)^2=2\log_ca\log_cb-\log_ca+\log_cb$$
You have a mixture of $\log_ca$ and $\log_cb$. Try converting everything to $\log_ca$ or to $\log_cb$, using the relation $b^2=ac$.

Since $a$, $b$, $c$ form a GP, you have square of middle term = product of the other two terms. This applies to any three consecutive terms of any GP.

Thus
$$b^2=ac\ \implies\ 2\log_cb=\log_ca+1.$$

Dhamnekar Winod

Active member
You have a mixture of $\log_ca$ and $\log_cb$. Try converting everything to $\log_ca$ or to $\log_cb$, using the relation $b^2=ac$.

Since $a$, $b$, $c$ form a GP, you have square of middle term = product of the other two terms. This applies to any three consecutive terms of any GP.

Thus
$$b^2=ac\ \implies\ 2\log_cb=\log_ca+1.$$

Hello,

Your answer is correct. $\frac32$ common difference of A.P is wrong.