f ( x ) = log x − 2 ⁡ ( 2 x + 3 7 − x ) &lt; 1Which is the solution of f ( x ) &lt; 1 ? I did: 2 x + 3 7 − x &gt; 0 x &lt; 7 results in x &gt; − 3 2 whereas x &gt; 7 results in x &lt; − 3 2 so x=( − 3 2 ,7) and also x − 2 &gt; 0 so x &gt; 2 x = ( 2 , 7 ) − { 3 }So after finding out where f is defined, I did this: log x − 2 ⁡ ( 2 x + 3 7 − x ) &lt; log x − 2 ( x − 2 ) 2 x + 3 7 − x &lt; x − 2 x 2 − 7 x + 17 &lt; 0And I figured I must have done a mistake somewhere as d &lt; 0 and x 2 's coefficient is &gt; 0 so f would be &gt; 0. Could you let me know where is my mistake? The correct answer is x = ( 2 , 3 )

Question

f  (  x  )  =      log          x      −      2        ⁡      (                  2        x        +        3                    7        −        x              )    &amp;lt;  1Which is the solution of      f  (  x  )  &amp;lt;  1 ? I did:               2      x      +      3              7      −      x        &amp;gt;  0     x  &amp;lt;  7 results in      x  &amp;gt;            −      3        2   whereas      x  &amp;gt;  7 results in      x  &amp;lt;            −      3        2   so x=(               −      3        2  ,7) and also      x  −  2  &amp;gt;  0 so      x  &amp;gt;  2     x  =  (  2  ,  7  )  −  {  3  }So after finding out where  f is defined, I did this:      log          x      −      2        ⁡      (                  2        x        +        3                    7        −        x              )    &amp;lt;  log                x      −      2            (    x    −    2    )                 2      x      +      3              7      −      x        &amp;lt;  x  −  2         x    2    −  7  x  +  17  &amp;lt;  0And I figured I must have done a mistake somewhere as      d  &amp;lt;  0 and          x    2  &#039;s coefficient is      &amp;gt;  0 so      f would be      &amp;gt;  0. Could you let me know where is my mistake? The correct answer is      x  =  (  2  ,  3  )

Emaidedip6g · Accepted Answer

log          (      x      −      2      )        ⁡      (                  2        x        +        3                    7        −        x              )    &amp;lt;  1Hint  1  )  x  −  2  &amp;gt;  0 and   x  −  2  ≠  1  2  )             2      x      +      3              7      −      x        &amp;gt;  0  2  )  If   x  −  2  &amp;gt;  1 then            2      x      +      3              7      −      x        &amp;lt;  x  −  2  3  ) If   0  &amp;lt;  x  −  2  &amp;lt;  1 then            2      x      +      3              7      −      x        &amp;gt;  x  −  2Now you can solve it.

tarjetaroja2t · Accepted Answer

The domain gives   2  &amp;lt;  x  &amp;lt;  7,   x  ≠  3 and since             2      x      +      3              7      −      x        ≠  x  −  2we get the answer without any cases immediately by the intervals method:  (  2  ,  3  )  .Done!

f(x)=log_(x-2)((2x+3)/(7-x))<1

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