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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is that the answer?
I don't know. I'm trying to figure this out.
ok 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 * (y-x/2) -- [log2 value comes form equation 3)
= -2y + x .. this is your final answer ..
I screwed up when i said the answer earlier ..
Wow. 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!
hahah .. I am a man buddy!! .. thanks for da medal ...