Question

Logarithm Question !!!

Answer 1

Shoot.

Answer 2

awesome!

Answer 3

\[\log_{a}3x^2 , \log_{a}3x , \log_{a}3 \] what is the common difference and the formula for the nth term

Answer 4

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)

Answer 5

what is log_(a)x * n equal to?

Answer 6

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.

Answer 7

log_ a x^n

Answer 8

how did you go from log3 +2logx + (n-1)(-logx) to log3 +(3-n)logx ?

Answer 9

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

Answer 10

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
\[\large{a_n = \log_a(3x^2)-(n-1)\log_a(x) = \log_a(3x^{3-n})}\]

Answer 11

yup

Answer 12

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...

Answer 13

If that was addressed to me, I just rewrote it in LaTeX so looked nicer. I know.

Answer 14

how did you go from here log3 +2logx + (1-n)(logx) to log3+(2+1-n) logx

Answer 15

take logx common...