Logarithms melting my brain So I've got an inequality: ln &#x2061;<!-- ⁡ --> ( 2 x

Aditya Erickson

Aditya Erickson

Answered question

2022-05-18

Logarithms melting my brain
So I've got an inequality: ln ( 2 x 5 ) > ln ( 7 2 x ) and I attempt to solve by doing the following:
ln ( 2 x ) ln ( 5 ) > ln ( 7 ) ln ( 2 x )
ln ( 2 x ) ln ( 2 x ) > ln ( 7 ) ln ( 5 )
ln 2 ( 2 x ) < ln ( 7 ) ln ( 5 )
2 x < ln ( 7 ) ln ( 5 ) e 2
I thought I could multiply by e 2 to get rid of ln 2 but I guess not...I thought they were inverses??? So anyways, my solution seems WAY off as the solution to the problem is: 3 < x < 7 2

Answer & Explanation

bshizzey7h

bshizzey7h

Beginner2022-05-19Added 9 answers

You applied an identity for logarithms in the wrong direction. What we have is
ln ( a / b ) = ln ( a ) ln ( b )
and not the other way around: ln ( a b ) ln ( a ) / ln ( b )
In fact, we don't have formula for ln ( a b )
But here what we only need to say is that the logarithm ln is a strictly increasing function throughout its domain (!!!), so for any numbers A , B in the domain of ln, we have ln ( A ) > ln ( B ) iff A > B   (and this is simply because X > Y     e X > e Y ).
So, now we then only need to solve 2 x 5   >   7 2 x, and care about only those values x for which both ln ( 2 x 5 ) and ln ( 7 2 x ) are defined.

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