anyone know anything about logarithms?

Question

my answer is log base 3 of 4 + log base 3 of (x+1) - 3log base 3 of (5+x)

is my answer totally complete??? or can i break it down more?

oh sorry i didnt explain fully i am supposed to express all exponents as factors so it started out as log base 3 (4(x+1)/(5+x)^3)

Answer 1

It looks expanded already. Are you trying to simplfy this? You can use the rules:

log[base b](a) + log[base b](c) = log[base b](ac)
log[base b](a) - log[base b](c) = log[base b](a/c)
log[base b](a^c) = c log[base b](a)

So in this case we have:
log[base 3](4) + log[base 3](x+1) - log[base 3]( (5+x)^3 ) =
log[base 3]( 4(x+1) ) - log[base 3]( (5+x)^3 ) =
log[base 3]( 4(x+1) / (5+x)^3 )

Answer 2

Lets see. First the base is irrelevant, so we'll ignore it. (It would become important if you need to caclulate a value, however)(it couldn't be ignored if there were more than one base, either) ( log base z of () = k * log base10 of () = c*ln() )

say z = log(4) + log(x+1) - 3log(5+x) so lets also say z=log(y)
so log(y) = log(4) + log(x+1) - 3log(5+x)
or y=4(x+1)/((5+x)^3) = (4x+4)/((x+1) + 4 )^3)
=4x+4/ ( (x+1)^3 +12(x+1)² + 48(x+1) + 64 )
or log(y)= z =
log(4)+log(x+1) - log((x+1)^3 +12(x+1)² + 48(x+1) + 64)
...Nope... I don't see any way to break it down further than what you've done.

Answer 3

Well, you could combine things. As long as the bases are the same, the log operations for log base 10 are valid for any other base. Thus you have 4(x+1)/ (5+x)^3

Answer 4

Mult logs.

log base 3 (4[x+1]) = log base 3 (4x + 4)