log 81^x - log 27^x = ?

Question

I have had three different answers

2x log 3
2x log 2
x log 2


what did you come up with?

Answer 1

xlog3

log81^x-log27^x=log(81^x / 27^x)=log(81`27)^x=log3^x=xlog3

Answer 2

log 81^x - log 27^x

{ i just remind you you can convert " - " to " / " , b coz we can work with it easier }

first i just need to factor ' log '
i just do it now ;
log 81^x - log 27^x = log (81^x - 27^x )

now i just convert - to /
log (81^x / 27^x ) , factor ' x ' ; log ( 81 / 27) ^ x = log (3)^x

well i think it would be log 3^x
any way good luck , hope u find the correct answer.

Answer 3

Solution:

log[(9^x)(9^x)]= 2log(9^x) = 2xlog(9)

log[(9*3)^x] = x(log(9) + log(3))

So, 2x(log(9) - x log(9) - x log(3)

= xlog(9) - x log(3)
= 2x log(3) - x log(3)
= x log(3) which is the correct answer