2log(base 6)12 - log(base 6)4?

Hi,

2log(base 6)12 - log(base 6)4 =

log(base 6)12² - log(base 6)4 =

log(base 6)144 - log(base 6)4 =


. . . . . . . . . . 144
log(base 6) = ------- = log(base 6)
. . . . . . . . . . . .4

log(base 6) 36 =

log(base 6) 6² = 2 <==ANSWER

I hope that helps!! :-)

2log_6(12) - log_6(4)
= log_6(12^2) - log_6(4)
= log_6(144/4)
= log_6(36)
= 2

Let log represent log to base 6 :-
2 log 12 - log 4
log 12² - log 4
log ( 12² / 4 )
log 36
2

this is how it goes.. =)
2 log (base 6) 12 - log (base 6) 4
= log (base 6) 144 - log (base 6) 4
= log (base 6) [144/4]
= log (base 6) 36 --> 6 x 6 = 36
= 2

hope it was of some help! :)

Nope. The answer is 2.

Since most calculators cannot evaluate log to any base other than ten, you have to rewrite the log expression in terms of logs written with the base of ten. Such formula for conversion is called CHANGE OF BASE FORMULA:

log M to the base N = (log M) / (log N)

Applying the rule to your question,
2log(base 6)12 - log(base 6)4 = 2 (log 12)/(log6) - (log4)/(log6)

Typing that into your calculator would give you 2.

log[6]144 - log[6]4

Remember this rule of logs --> log a - log b = log(a/b)

log[6](144/4)
log[6](36)

Rewrite that as 6²:
log[6](6²)
2 log[6](6)

Rule log[b](b) = 1
2 * 1
= 2

2log(base 6)12 − log(base 6)4
=log(base 6)12² − log(base 6)4
=log(base 6){12²/4}
=log(base 6){36}
=log(base 6){6²} =2log(base 6){6} =2×1=2

log(base 6)12^2 - log(base 6)4 = log(base 6) (144/4) = log(base 6) 36
= log(base 6) 6^2 = 2

2

log(base 6)12^2 - log(base)4 = log(base 6)(144/4) = log(base 6)36 = log(base 6)(6*6) = 2