# Thread: summation-related(?) problem

1. ## summation-related(?) problem

Hello,

I'm not sure if this is the right forum -- I'm not a student so I've not arrived at this problem in a textbook context.

I want to solve an equation of the form:

(a^n)b-(a^n-1)c-(a^n-2)c-...-ac-c=0

where a, b, and c are constants and n is the unknown to be found.

Any pointers -- whether on the problem or on where to go to pose it again -- would be appreciated.

Regards,
Isolde.

2. Originally Posted by isolde
Hello,

I'm not sure if this is the right forum -- I'm not a student so I've not arrived at this problem in a textbook context.

I want to solve an equation of the form:

(a^n)b-(a^n-1)c-(a^n-2)c-...-ac-c=0

where a, b, and c are constants and n is the unknown to be found.

Any pointers -- whether on the problem or on where to go to post it again -- would be appreciated.

....Double Posting is against rules if you want to move your thread report it to mods, anyway you have posted it correctly

Regards,
Isolde.
$(a^n)b-(a^{n-1})c-(a^{n-2})c-...-ac-c=0$

$a^{n}b = (a^{n-1})c+(a^{n-2})c...+ac+c$

$a^{n}b = c[(a^{n-1})+(a^{n-2})...+a+1]$

**Now $[(a^{n-1})+(a^{n-2})...+a+1]$ is a Geometric Progression

**It is $= 1 \frac{a^{n} - 1}{a-1}$

$a^n b = c\times \frac{a^{n} - 1}{a-1}$

$a^{n+1}b - a^n b = c(a^n - 1)$

Put $a^n =t$

$abt +c = t(b+c)$

$t(b+c-ab) = c$

Hence
$
t= \frac{c}{b+c-ab}$

$
a^n = \frac{c}{b+c-ab}$

Hence

$n \ln a = \ln \frac{c}{b+c-ab}$

$\implies n = \ln (\frac{c}{ab+ac-a^2b})$

3. Thanks for the solution ADARSH.

4. Im new to logarithms, so its taken me a while to catch on to what youre doing here. I still dont understand the last step.

From

$
n \ln a = \ln \frac{c}{b+c-ab}
$

I assume you go by the intermediate step

$
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}
$

in order to derive

$
n = \ln (\frac{c}{ab+ac-a^2b})
$

but as I say, not knowing logs very well, I cant see how this last step to the solution is made. Presumably by some such identity as

$
\frac{\log_a{b}}{\log_a{c}} = \log_a{\frac{b}{c}}
$

but Im not familiar with this one, and havent been able to find it at Wikipedias list.

I have another thought. Would it be possible from

$
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}
$

to derive

$
n = \log_a{\frac{c}{b+c-ab}}
$

by the identity

$
\log_a{b} = \frac{\log_c{b}}{\log_b{c}}
$

?

Thanks again,
Marc.

5. Originally Posted by isolde
Im new to logarithms, so its taken me a while to catch on to what youre doing here. I still dont understand the last step.

From

$
n \ln a = \ln \frac{c}{b+c-ab}
$

I assume you go by the intermediate step

$
n = \frac{\ln \frac{c}{b+c-ab}}{\ln a}
$

in order to derive

$
n = \ln_a {(\frac{c}{b+c-ab})}
$

--------------------------------------------------------------
it was wrong I have corrected it
You are correct
You won't find it because it's wrong, Sorry My mistake

6. Thank-you for your gracious response.