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|dw:1360922596470:dw|

correct so far?

you got the 1st and second term correct.
whats your 3rd term ? check/try it again

log 2(6) + log 2(2^1/2)

Also, log of 1 in any base = 0
because a^0 =1

yea so it will be 1/2

well the 3rd term

so -log2(6) - 1/2?
:D

correct, so are you clear till here :
\(0-(\log 6+1/2)\)
now how can you write 6 as ?

yes, but its still needs to simplify

oh you can write 6 as
-(log2(3) + log2(2) +1/2)
-(log2(3) + 3/2)

-log2(3) - 3/2
:D

very good!
\(\huge \color{red}{\checkmark }\)

✓

hehe mine sucks lol

copy and paste doesnt work :(

or copy this :
`\huge \color{red}{\checkmark }`
and write it in between
`\[` and `\]`

you can change the color if you want :)

Thanks :D

nice :) welcome ^_^