Here's the question you clicked on:
aceace
Logarithm Question !!!
\[\log_{a}3x^2 , \log_{a}3x , \log_{a}3 \] what is the common difference and the formula for the nth term
log 3x-log 3x^2 = log 3 - log 3x = - log x = d =common difference S= n/2 (2 a1+(n-1)d) = n/2 (2 log 3x^2 - (n-1)log x)
what is log_(a)x * n equal to?
common difference is -log x (base as a ) nth term is given by log3 +2logx + (n-1)(-logx)= log3 +(3-n)logx (throughout base is a as given in the question.
how did you go from log3 +2logx + (n-1)(-logx) to log3 +(3-n)logx ?
log3 +2logx + (n-1)(-logx) = log3 +2logx + (1-n)(logx)= log3+(2+1-n) logx =log3+(3-n)logx= log3x^(3-n) take base as a
This is making my eyes bleed. \[ \log_a(3x^2)-\log_a(x) = \log_a(3x) \text{ etc...} \] so if \[a_1 = \log_a(3x^2) \] we have \[\large{a_n = \log_a(3x^2)-(n-1)\log_a(x) = \log_a(3x^{3-n})}\]
if n=1 it gives the first term u gave and for n=2 it gives log3x ur second term if n=3 it gives third term as log as mentioned by u...
If that was addressed to me, I just rewrote it in LaTeX so looked nicer. I know.
how did you go from here log3 +2logx + (1-n)(logx) to log3+(2+1-n) logx