# PreCalculus

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Hi i don't know how to do this problem

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 -

let 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 -

hmmm
how did you go from this
a(log3+logn)=b(log4+logn)
to this?
logn(a-b)=blog4-alog3
logn=( )/(a-b)

• PreCalculus -

You are supposed to fill in the blanks.

alog3+alogn=blon4=blogn
logn(a-b)=blog4-alog3
logn= (blog4-alog3)/(a-b)

• RESPONSE -

well 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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