# math

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

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3. 👁 107
1. 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.

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posted by drwls
2. 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

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posted by Tarun
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 log3[64 ÷ 8/3 x 2]
= log3[64 x 3/8 x 2]
= log3 48

etc

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posted by Reiny
4. 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.....?

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posted by Tarun
5. ..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)
= log3 (2 ÷ 8/3)
= log3 (2 x 3/8)
= log3 (6/8)

now you have log3(64 x 6/8)
= log348

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posted by Reiny
6. oo I c it...my bad. Thnx 4 explain reiny! ._.

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posted by Tarun
7. sorry, did not mean to print all that in bold,
forgot to turn it off after NOT

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posted by Reiny
8. haha its aight......shud have used instead of </b? :D

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posted by Tarun

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