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aceace
 2 years ago
Best ResponseYou've already chosen the best response.0\[\log_{a}3x^2 , \log_{a}3x , \log_{a}3 \] what is the common difference and the formula for the nth term

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0log 3xlog 3x^2 = log 3  log 3x =  log x = d =common difference S= n/2 (2 a1+(n1)d) = n/2 (2 log 3x^2  (n1)log x)

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0what is log_(a)x * n equal to?

matricked
 2 years ago
Best ResponseYou've already chosen the best response.0common difference is log x (base as a ) nth term is given by log3 +2logx + (n1)(logx)= log3 +(3n)logx (throughout base is a as given in the question.

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0how did you go from log3 +2logx + (n1)(logx) to log3 +(3n)logx ?

matricked
 2 years ago
Best ResponseYou've already chosen the best response.0log3 +2logx + (n1)(logx) = log3 +2logx + (1n)(logx)= log3+(2+1n) logx =log3+(3n)logx= log3x^(3n) take base as a

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.0This 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 \[\large{a_n = \log_a(3x^2)(n1)\log_a(x) = \log_a(3x^{3n})}\]

matricked
 2 years ago
Best ResponseYou've already chosen the best response.0if 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...

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.0If that was addressed to me, I just rewrote it in LaTeX so looked nicer. I know.

aceace
 2 years ago
Best ResponseYou've already chosen the best response.0how did you go from here log3 +2logx + (1n)(logx) to log3+(2+1n) logx
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