Solve ( l o g 2 ( x + 1 ) ) 2 = 4 ( l o g 2 ( x + 1 ) ) 2 = 4 l o g 2 ( x + 1 ) ∗ l o g 2 ( x + 1 ) = l o g 2 16 x 2 + 2 x − 15 = 0 ( x + 1 ) ∗ ( x + 1 ) = 16 x 2 + 2 x + 1 = 16 x 2 + 2 x − 15 = 0 ( x + 5 ) ( x − 3 ) = 0 x 1 = − 5 ; x 2 = 3The solution is only x 1 = 3. Is this correct?

Question

Solve   (  l  o      g          2        (  x  +  1  )      )    2    =  4  (  l  o      g          2        (  x  +  1  )      )    2    =  4  l  o      g          2        (  x  +  1  )  ∗  l  o      g          2        (  x  +  1  )  =  l  o      g          2        16      x          2        +  2  x  −  15  =  0  (  x  +  1  )  ∗  (  x  +  1  )  =  16      x          2        +  2  x  +  1  =  16      x          2        +  2  x  −  15  =  0  (  x  +  5  )  (  x  −  3  )  =  0      x    1    =  −  5  ;      x    2    =  3The solution is only       x    1    =  3. Is this correct?

Kaydence Washington · Accepted Answer

If  (      log    2    ⁡  (  x  +  1  )      )    2    =  4then      log    2    ⁡  (  x  +  1  )  =      +    2  or      log    2    ⁡  (  x  +  1  )  =      −    2  so   x  +  1  =      2                  +        2             or   x  +  1  =      2                  −        2

Lydia Carey · Accepted Answer

Your solution is incorrect. First note that      log    2    ⁡      (          a      2        )    ≠            (              log        2            ⁡      (      a      )      )        2  In your case, we have            (              log        2            ⁡      (      x      +      1      )      )        2    =  4    ⟹        log    2    ⁡  (  x  +  1  )  =  ±  2    ⟹    x  +  1  =      2          ±      2          ⟹    x  +  1  =  4   or   x  +  1  =            1      4      Hence,  x  =  3   or   x  =  −            3      4

Solve ( l o g 2 ( x + 1 ) ) 2 = 4...

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