Logarithm problem : Prove that l o g <mrow class="MJX-TeXAtom-ORD">

migongoniwt 2022-06-27 Answered
Logarithm problem : Prove that l o g 3 2 1 2 > 0
My approach :
l o g 3 2 1 2 > 0
1 2 l o g 3 1 2 > 0
1 2 [ l o g 3 1 l o g 3 2 ] > 0
1 2 [ 0 l o g 3 2 ] > 0
1 2 [ l o g 3 2 ] > 0 { which is false}
Please suggest... thanks...
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Answers (2)

g2joey15
Answered 2022-06-28 Author has 21 answers
It is hard to prove something that isn't true. Alpha shows log 9 1 2 < 0.3 so you aren't even close. Generally, logs of things less than 1 are negative.
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pokoljitef2
Answered 2022-06-29 Author has 9 answers
To evaluate l o g 3 2 1 2 you solve ( 3 2 ) x = 1 2 x l o g ( 3 2 ) = l o g ( 1 2 ) x = l o g ( 1 2 ) l o g ( 3 2 ) , but l o g ( 1 2 ) = l n ( 2 ). Hence, l o g 3 2 1 2 is not greater than 0.
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