math
posted by Jz .
i have 2 math questions i don't understand so if you could help me i would like that.
Log base 3 64log base 3 (8/3)+log base 3 2=log base 3 4r
Log base 6 (b^2+2)+logbase6 2=2

Rewrite the first as
Log(3) [64*2/(8/3)] = Log(3) 4r
If the logs to the same base of two numbers are the same, then the numbers themselves must be the same.
Therefore 128*3/8 = 4r
48 = 4r
r = 12
================
Log(6)[(b^2+2)*2] = Log(6)36
See what I did? I rewrote 2 as log(6)36
2 (b^2 + 2) = 36
b^2 + 2 = 18
Take it from there. There are two answers. 
Hey I think u got the 2nd question wrong drwls...wudnt it be
Log 3 64Log 3 (8/3) + Log 3 2 = Log 4 r
Log 3 64  Log 3 (16/3) = Log 4r
Log 3 (63*3/16) = Log 4r
Log 3 12 = Log 4r
12=4r
r=3 
Nope, drwls is correct, I got the same result as he did
remember that multiplication and division are done in the order they come
so the left side is log_{3}[64 ÷ 8/3 x 2]
= log_{3}[64 x 3/8 x 2]
= log_{3} 48
etc 
but in the original question was
Log 3 64Log 3 (8/3) + Log 3 2 = Log 4 r
so, wudnt u do the addition first and then do the subtraction ... in other words solve Log 3 (8/3) + Log 3 2 and then do the rest.....? 
..but it's NOT</b? Log 3 (8/3) + Log 3 2
it is Log 3 (8/3) + Log 3 2 or
Log 3 2  Log 3 (8/3)
= log_{3} (2 ÷ 8/3)
= log_{3} (2 x 3/8)
= log_{3} (6/8)
now you have log_{3}(64 x 6/8)
= log_{3}48 
oo I c it...my bad. Thnx 4 explain reiny! ._.

sorry, did not mean to print all that in bold,
forgot to turn it off after NOT 
haha its aight......shud have used instead of </b? :D
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