Asked by Jz
                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
            
            
        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
Answers
                    Answered by
            drwls
            
    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.
    
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.
                    Answered by
            Tarun
            
    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
    
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
                    Answered by
            Reiny
            
    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<sub>3</sub>[64 ÷ 8/3 x 2]
= log<sub>3</sub>[64 x 3/8 x 2]
= log<sub>3</sub> 48
etc
    
remember that multiplication and division are done in the order they come
so the left side is log<sub>3</sub>[64 ÷ 8/3 x 2]
= log<sub>3</sub>[64 x 3/8 x 2]
= log<sub>3</sub> 48
etc
                    Answered by
            Tarun
            
    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.....?
    
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.....?
                    Answered by
            Reiny
            
    ..but it's <b>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<sub>3</sub> (2 ÷ 8/3)
= log<sub>3</sub> (2 x 3/8)
= log<sub>3</sub> (6/8)
now you have log<sub>3</sub>(64 x 6/8)
= log<sub>3</sub>48</b>
    
it is -Log 3 (8/3) + Log 3 2 or
Log 3 2 - Log 3 (8/3)
= log<sub>3</sub> (2 ÷ 8/3)
= log<sub>3</sub> (2 x 3/8)
= log<sub>3</sub> (6/8)
now you have log<sub>3</sub>(64 x 6/8)
= log<sub>3</sub>48</b>
                    Answered by
            Tarun
            
    oo I c it...my bad. Thnx 4 explain reiny! ._.
    
                    Answered by
            Reiny
            
    sorry, did not mean to print all that in bold,
forgot to turn it off after <b>NOT</b>
    
forgot to turn it off after <b>NOT</b>
                    Answered by
            Tarun
            
    haha its aight......shud have used </b> instead of </b? :D
    
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