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Prove. 3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))Asked by Hasane
Prove:
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
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Answered by
Steve
3/(log_2 (a)) - 2/(log_4 (a)) = 1/(log_(1/2)(a))
since 4 = 2^2, log_4(a) = 1/2 log_2(a)
since 1/2 = 2^-1, log_(1/2)(a) = -log_2(a)
so, if we let x = log_2(a), we have
3/x - 2/(x/2) = 1/(-x)
3/x - 4/x = -1/x
(4-3)/x = -1/x
1/x = -1/x
???
Is there a typo somewhere ?
since 4 = 2^2, log_4(a) = 1/2 log_2(a)
since 1/2 = 2^-1, log_(1/2)(a) = -log_2(a)
so, if we let x = log_2(a), we have
3/x - 2/(x/2) = 1/(-x)
3/x - 4/x = -1/x
(4-3)/x = -1/x
1/x = -1/x
???
Is there a typo somewhere ?
Answered by
Hasane
i.imgur[dot]com/hE0sWBt[dot]gif
Answered by
Hasane
Make sure its just like that, it has to be solvable o.o
Answered by
Hasane
er "prove-able"
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