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anonymous
 5 years ago
given log6(4)=X, log6(10)=y, log6(9)=z then find log6(1/25) in terms of X, Y & Z
anonymous
 5 years ago
given log6(4)=X, log6(10)=y, log6(9)=z then find log6(1/25) in terms of X, Y & Z

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the answer 2Y+\[\sqrt{?}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean 2y + squareroot(x)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How did you get that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't know. I'm trying to figure this out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok here is how i did it: I am not writing base since it is 6. I will just write numbers involved.. log(1/25) = log(5^2) = 2*log5  eqn (i) [so log5 is what you are looking for] log10 = log(2*5) = log2 + log5 = y log5 = y  log2  eqn (ii) .. [so you will need value of log2 to solve log5] log4 = log(2^2) = 2*log2 = x log2 = x/2  equation(iii) finally combining all back to equation i log(1/25) = 2*log5 = 2 * ( y  log2) [from equation ii put value for log 5) =2 * (yx/2)  [log2 value comes form equation 3) = 2y + x .. this is your final answer .. I screwed up when i said the answer earlier ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wow. That's great. I'll try to work through it to make sure I understand it. But I think I got it. Thanks! U da man. Unless you're a woman. Either way, thanks again!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hahah .. I am a man buddy!! .. thanks for da medal ...
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