# 2 more math questions

Discussion in 'DAT Discussions' started by ATLATLATL, May 2, 2007.

1. ### ATLATLATL 2+ Year Member

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if log2=A and log3=B then log six is?
a+b
a-b
a*b
a/b
a^b

i would've thought a*b

oh and the other question is...

with one fair die, find the probability of throwing two fours in five attempts?

2. ### gochi 10+ Year Member

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I cant really explain this, but I know how it works, so let me illustrate.

Log X + Log Y= Log(XY) likewise Log X - Log Y= Log(X/Y)

These are just properties of LOGARITHMS.

Q: if log2=A and log3=B then log six is?

A: log 6= log2 + log3=log(2*3)=log6 which is the same as a=3;b=3--> ab.

3. ### Lonely Sol cowgoesmoo fan! 10+ Year Member

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Would this a binomial probablity because probability of sucess equals 1-probability oif failure. The reason is because there is only two choices; rolling a 4 or not rolling a 4:

Since you have 1/6 chances of throwing 4 and 5/6 chance of throwing anything but 4.

Since, you want 2 4s in 5 attenpts, you do the following:

(1/6*1/6*5/6*5/6*5/6 )*10 (I got 10 from 5C2)

*plz correct me if I am wrong

4. ### Streetwolf Ultra Senior Member Dentist 7+ Year Member

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Want two 4s in 5 tries.

5 choose 2 - picks the two throws that will be a 4.
(1/6)^2 - odds of throwing two 4s.
(5/6)^3 - odds of throwing three non-4s.

10*(1/36)*(125/216) = 1250/7776 = (625/3888) = 0.16075

So 16.075%.

Correct, LS.

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5. ### Streetwolf Ultra Senior Member Dentist 7+ Year Member

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= 10^(Log X) * 10^(Log Y) <-- 10^(Log X) = Log(10^X) = X
= 10^(Log X + Log Y) <-- A^X * A^Y = A^(X+Y)

So we have XY = 10^(Log X + Log Y).
Take the Log of both sides:

Log(XY) = Log(10^(Log X + Log Y)
Log(XY) = Log X + Log Y

Try it for Log(X/Y).

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