Posted by **Jz** on Sunday, February 10, 2008 at 8:01pm.

i have 2 math questions i don't understand so if you could help me i would like that.

Log base 3 64-log base 3 (8/3)+log base 3 2=log base 3 4r

Log base 6 (b^2+2)+logbase6 2=2

- math -
**drwls**, Sunday, February 10, 2008 at 8:18pm
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.

- math -
**Tarun**, Sunday, February 10, 2008 at 8:38pm
Hey I think u got the 2nd question wrong drwls...wudnt it be

Log 3 64-Log 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

- math -
**Reiny**, Sunday, February 10, 2008 at 8:57pm
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

- math -
**Tarun**, Sunday, February 10, 2008 at 9:00pm
but in the original question was

Log 3 64-Log 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.....?

- math -
**Reiny**, Sunday, February 10, 2008 at 9:36pm
..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

- math -
**Tarun**, Sunday, February 10, 2008 at 9:38pm
oo I c it...my bad. Thnx 4 explain reiny! ._.

- math -
**Reiny**, Sunday, February 10, 2008 at 9:38pm
sorry, did not mean to print all that in bold,

forgot to turn it off after **NOT**

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**Tarun**, Sunday, February 10, 2008 at 9:47pm
haha its aight......shud have used instead of </b? :D

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