How prove ( 2 + 5 x ) ln ⁡ x − 6 ( x − 1 ) &gt; 0. ∀ x &gt; 1let x &gt; 1 show that: ( 2 + 5 x ) ln ⁡ x − 6 ( x − 1 ) &gt; 0. ∀ x &gt; 1Let f ( x ) = ( 2 + 5 x ) ln ⁡ x − 6 ( x − 1 ) , f ′ ( x ) = 2 x + 5 ln ⁡ x − 1since f ( 1 ) = 1.so it must prove f ′ ( x ) = 2 x + 5 ln ⁡ x − 1 &gt; 0 ?

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How prove   (  2  +  5  x  )  ln  ⁡      x    −  6  (  x  −  1  )  &amp;gt;  0.  ∀  x  &amp;gt;  1let   x  &amp;gt;  1 show that:  (  2  +  5  x  )  ln  ⁡      x    −  6  (  x  −  1  )  &amp;gt;  0.  ∀  x  &amp;gt;  1Let  f  (  x  )  =  (  2  +  5  x  )  ln  ⁡      x    −  6  (  x  −  1  )  ,               f    ′    (  x  )  =            2      x        +  5  ln  ⁡      x    −  1since   f  (  1  )  =  1.so it must prove      f    ′    (  x  )  =            2      x        +  5  ln  ⁡      x    −  1  &amp;gt;  0  ?

Kody Larsen · Accepted Answer

Let   f  (  x  )  =  ln  ⁡      x    −            6      (      x      −      1      )              5      x      +      2      Hence,      f    ′    (  x  )  =      1    x    −  6  ⋅            5      x      +      2      −      5      (      x      −      1      )              (      5      x      +      2              )        2              =      1    x    −      42          (      5      x      +      2              )        2              =  =            25              x        2            −      22      x      +      4              x      (      5      x      +      2              )        2              =            25              x        2            −      25      x      +      3      x      +      4              x      (      5      x      +      2              )        2              &amp;gt;  0for all   x  &amp;gt;  1Thus,   f  (  x  )  &amp;gt;  f  (  1  )  =  0 and we are done!

crazygbyo · Accepted Answer

write your inequality for  x  &amp;gt;  1in the form  ln  ⁡  (  x  )  −            6      (      x      −      1      )              2      +      5      x        &amp;gt;  0and define  f  (  x  )  =  ln  ⁡  (  x  )  −            6      (      x      −      1      )              2      +      5      x      and compute the derivative with respect to   x

How prove (2+5x)lnx−6(x−1)>0.AAx>1

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