Education Help with logarithms

Im stuck on a question on logs can someone help me out.
I need to express log(2 sqrt{10}) - 1/3 log 0.8 - log (10/3) in the form c + log d where c and d are rational number and logs are to base 10.

What I did:
2 sqrt{10} = sqrt{40} = 40^{1/2}
So log(2 sqrt{10}) = 1/2 log 40

log (10/3) = log 10 - log 3
So -log(10/3) = -log 10 + log 3

-log 10 = -1

SO: so far ive simplified to: 1/2 log 40 - 1/3 log 0.8 - 1 + log 3

Is this right so far?
If so what do i do next

 

When adding or subtracting logs you can divide or multiply. For example Ln(5) - Ln(4) = Ln(5/4).

 

Yes I know but i believe when they have the same number before them. For example, 2 log (5) - log (4) cannot be worked out that way. (I think. Correct me if i am wrong)

 

2 log (5) - log (4) = log (5²) - log (4) = log(5²/4)

 

Can't believe i didn't think of that.
Thanks guys. And does that mean what I've done so far is ok?

 

2 log (5) would be the same as log (5)^2

log(2 sqrt{10}) - 1/3 log 0.8 - log (10/3)

= Log (2) + 1/2 log (10) - log (0.8)^1/3 - [log (10) - log (3)]

= Log (2) + (0.5) - (log (0.8)^1/3) - [1-log (3)]
-0.5 + log (2) - (log (0.8)^1/3) + (log 3) )

thats as far as i can get in my head

-0.5 + log (2) - [ (log (0.8)^1/3) x 3) ]

-0.5 + [( log (2) ) ] / [ (log (0.8)^1/3) x 3) ]

try that? can you simplify that any further?

 

bixie thats as far as I can get. Anybody go any ideas?

 

log₁₀(10) = 1, so that's where your constant c will come from. Just separate out the 10s and combine the rest into the log(d).

I get -0.5 + log(6/(0.8^(1/3))).

If you want the working:

Spoiler

log (2√10) - (1/3)log(0.8) - log(10/3)
= log 2 + 0.5log(10) - (1/3)log(0.8) - log(10) + log(3)
= log 2 + 0.5 - (1/3)log(0.8) - 1 + log(3)
= -0.5 + log(6) - (1/3)log(0.8)
= -0.5 + log(6/(0.8^(1/3)))