wykaż, że
(1/ log₂π)+(1/log₅π)>2
L=(1/ log₂π)+(1/log₅π)
z zamiany podstaw mamy:
L=(log2/logπ)+(log5/logπ)=log10/logπ=1/logπ
1/logπ>2
1>2logπ ⇒logπ≈0.4971
1>0.9942
cnd.
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L=(1/ log₂π)+(1/log₅π)
z zamiany podstaw mamy:
L=(log2/logπ)+(log5/logπ)=log10/logπ=1/logπ
1/logπ>2
1>2logπ ⇒logπ≈0.4971
1>0.9942
cnd.