log [base 3]y² = 2 log [base 3] y
(Log [base 3] y)²
equal to ?
they are different, rite!?
2007-05-05 02:28:52 · 7 answers · asked by ziping 1 in Science & Mathematics ➔ Mathematics
(log₃ y)²
and
log₃ (y)²
are different!
but (log₃ y)² equals to wad!?
XD
2007-05-05 02:57:04 · update #1
(log₃ y)²
and
log₃ (y)²
are different!
but (log₃ y)² equals to wad!?
XD
2007-05-05 02:59:09 · update #2
You are quite correct:
(log₃ y)² = (log₃ y)(log₃ y)
But, as you correctly state, log₃ y² = 2 log₃ y , and this is
NOT equal to (log₃ y)(log₃ y).
So they ARE different.
(log₃ y)² means (log₃ y)(log₃ y).
There is NO "formula" or identity for (log₃ y)².
It simply says: "take log to the base 3 of y, then square your answer." There isn't an alternative way to work it out.
2007-05-05 02:38:06 · answer #1 · answered by sumzrfun 3 · 0⤊ 1⤋
No, they are the same.Because it is a log rule.I'm giving you general formula for this rule:
log (base n) a*b = b log (base n) a
just like:
log (base 3) y*2 = 2 log (base 3) y.
This means that whenever the unknown has a power the power must come before the log.
Bye n hope that this is going to help u.
2007-05-05 10:04:48 · answer #2 · answered by hans 1 · 0⤊ 0⤋
No, they are equal!
log y² = 2*log y
regardless of the base.
Why?
log y² = log(y*y) = log y + log y = 2 log y.
2007-05-05 09:43:08 · answer #3 · answered by steiner1745 7 · 0⤊ 0⤋
no, they are the same
that is a log rule
log x^n = n ln x
2007-05-05 09:31:54 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I give the same answer as that of sumzrfun
2007-05-05 10:41:43 · answer #5 · answered by chapani himanshu v 2 · 0⤊ 0⤋
Use the Sanio calculator to solve your question.
2007-05-05 09:31:56 · answer #6 · answered by Hema 2 · 0⤊ 1⤋
.
(log [base3] y )^2 =(log[base3]y) (log[base3]y),,,so
.
.................................=(log[base3]y^2)/2 (log[base33]y^2 /2 =
.
....=(1/4) (log[base3]y^2.........................,
.
2007-05-05 10:01:46 · answer #7 · answered by Tuncay U 6 · 0⤊ 0⤋