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sarah12345
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express

log(2 rt 10) - (1/3)log 0.8 - log (10/3) in the form c+ log d where c and d are rational numbers


grrrrrrrrrrrrrrrrrrrrrrr cant do it can anyone help

thanks alot
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nas7232
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use the log rules from c2

log(2 rt 10) - (1/3)log 0.8 - log (10/3) in the form c+ log d where c and d are rational numbers

firstly simplify

log2(2 x 10^1/2) - log0.8^1/3 - log10 - log 3

Therefore:
1 + 1/2log 10 - log0.2 - log10 - log3

simplify = answer
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madhapper
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log(2 rt 10) - log (4/5) - log (10/3)=
log 2 + log(rt10) + log(5/4) + log(10/3)=
log(2.5/4.10/3) + 1/2log10=
log25/3+1/2
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evariste
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(Original post by sarah12345)
express

log(2 rt 10) - (1/3)log 0.8 - log (10/3) in the form c+ log d where c and d are rational numbers


grrrrrrrrrrrrrrrrrrrrrrr cant do it can anyone help

thanks alot
log(2 rt 10) - (1/3)log 0.8 - log (10/3)
=log2+log(rt(10))-1/3log(8/10)-{log(10)-log(3)}
=log(2)+1/2log(10)-1/3(log8-log(10))-1+log(3)
=log(2)+1/2-1/3log(2^3)+1/3-1+log(3)
=log(2)-log(2)+log(3)-1/6
=log(3)-1/6
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sarah12345
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(Original post by evariste)
log(2 rt 10) - (1/3)log 0.8 - log (10/3)
=log2+log(rt(10))-1/3log(8/10)-{log(10)-log(3)}
=log(2)+1/2log(10)-1/3(log8-log(10))-1+log(3)
=log(2)+1/2-1/3log(2^3)+1/3-1+log(3)
=log(2)-log(2)+log(3)-1/6
=log(3)-1/6

thankyou soooo much!
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