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April 20, 2014

April 20, 2014

Posted by **Kate** on Monday, April 26, 2010 at 8:08pm.

solve for n^3

given that log(40 SQRT(3))/log(4n) = log(45)/log(3n)

Thanks i'm copletely lost and don't know how to solve

- PreCalculus -
**bobpursley**, Monday, April 26, 2010 at 8:17pmlet a=log(40 sqrt(3)) and b=log45

then

a/log(4n)=b/log(3n)

a(log3n)=b*log(4n)

a(log3+logn)=b(log4+logn)

logn(a-b)=blog4-alog3

logn=( )/(a-b)

logn^3=3logn= 3*above

then take the antilog of both sides, it looks like fun.

- RESPONSE -
**RESPONSE**, Monday, April 26, 2010 at 8:21pmhmmm

how did you go from this

a(log3+logn)=b(log4+logn)

to this?

logn(a-b)=blog4-alog3

logn=( )/(a-b)

- PreCalculus -
**bobpursley**, Monday, April 26, 2010 at 8:27pmYou are supposed to fill in the blanks.

alog3+alogn=blon4=blogn

logn(a-b)=blog4-alog3

logn= (blog4-alog3)/(a-b)

- RESPONSE -
**RESPONSE**, Monday, April 26, 2010 at 8:49pmwell if follow now but don't know how to take the anti log of that mumbo jumbo

n = 10^(3((blog4-alog3)/(a-b) )

im unsure how that simplifies

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